**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make It Equal
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.

You can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \bmod n) + 1}$$$ by $$$1$$$.

After you perform the operations, all elements of the array should be non-negative equal integers.

Your task is to calculate the minimum number of operations you have to perform.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.

Examples:
input:
3
2
1 1
3
1 3 2
4
2 1 2 6
output:
0
-1
3

Note:

Buggy Code:
import java.io.*;
import java.util.*;

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = true;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static int n;
    static long sum, record;
    static int[] a;
    static void solve(PrintWriter o) {
        try {
            n = nextInt();
            a = new int[n];
            sum = 0l;
            record = 0l;
            for(int i=0;i<n;i++) {
                a[i] = nextInt(); // 1e9
                sum += a[i];
            }
            int l = 0, r = (int)1e9+1;
            while(l < r) {
                int m = (l+r)>>1;
                if(check(m)) l = m+1;
                else r = m;
            }
            if(l == 0) {
                o.println(-1);
                return;
            }
            l--;
            check(l);
            o.println(record);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    static boolean check(int m) {
        int[] b = a.clone();
        long remain = sum-1l*m*n, use = 0l;
        if(remain < 0) return false;
        for(int i=0;i<4*n;i++) {
            long d = b[i%n]-m;
            if(d > 0) {
                long t = d/2;
                int md = (int)(d%2);
                use += t;
                b[i%n] = m+md;
                b[(i+1)%n] += t;
            }
            if(use > remain) {
                return false;
            }
        }
        for(int i=0;i<n-1;i++) if(b[i] != b[i+1]) return false;
        use += 1l*(b[0]-m)*n;
        record = use;
        return use <= remain;
    }
    public static int upper_bound(long[] preSum, long val){
        int l = 0, r = preSum.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(preSum[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(List<long[]> li, long val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid)[0] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    static class Pair {
        Integer p1, p2;
        public Pair(Integer p1, Integer p2) {
            this.p1 = p1;
            this.p2 = p2;
        }
        @Override
        public int hashCode() {
            int prime = 31, ret = 0;
            ret = ret*prime+p1.hashCode();
            ret = ret*prime+p2.hashCode();
            return ret;
        }
        @Override
        public boolean equals(Object o) {
            if(o instanceof Pair pair) {
                return p1.equals(pair.p1) && p2.equals(pair.p2);
            }
            return false;
        }
    }
    static class DSU {
        int[] parent;
        int[] size;
        public DSU(int n) {
            parent = new int[n];
            size = new int[n];
            for(int i=0;i<n;i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }
        public int find(int x) {
            if(parent[x] != x) parent[x] = find(parent[x]);
            return parent[x];
        }
        public void union(int x, int y) {
            int root_x = find(x);
            int root_y = find(y);
            if(root_x == root_y) return;
            parent[root_x] = root_y;
            size[root_y] += size[root_x];
            size[root_x] = 0;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. Here are the steps to think through this:

1. **Edge Cases for Input Constraints:**
   - Test with the minimum and maximum values for `n` and `a_i`.
   - Test with all elements being the same.
   - Test with elements that are very close to each other but not exactly the same.

2. **Specific Scenarios:**
   - Test with an array where the sum of elements is not divisible by `n`.
   - Test with an array where the elements are already equal.
   - Test with an array where it is impossible to make all elements equal.

3. **Small-Scale Input:**
   - Ensure the input is small enough to be easily validated and fits within the constraints.

Based on these considerations, let's construct a test input:

- **Test Case 1:** All elements are equal.
- **Test Case 2:** The sum of elements is not divisible by `n`.
- **Test Case 3:** It is impossible to make all elements equal.

Here is the test input:

```plaintext
3
3
1 1 1
4
2 1 2 6
5
1 2 3 4 5
```

Explanation:
1. **Test Case 1:** All elements are equal (1, 1, 1). The expected output is 0 because no operations are needed.
2. **Test Case 2:** The sum of elements (2 + 1 + 2 + 6 = 11) is not divisible by 4. The expected output is -1 because it is impossible to make all elements equal.
3. **Test Case 3:** The elements are (1, 2, 3, 4, 5). This is a more complex case where the elements are close but not equal. The expected output is a positive number of operations, which the algorithm should correctly calculate.

This test input should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make It Equal
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given an integer array $$$a$$$ of size $$$n$$$. The elements of the array are numbered from $$$1$$$ to $$$n$$$.

You can perform the following operation any number of times (possibly, zero): choose an index $$$i$$$ from $$$1$$$ to $$$n$$$; decrease $$$a_i$$$ by $$$2$$$ and increase $$$a_{(i \bmod n) + 1}$$$ by $$$1$$$.

After you perform the operations, all elements of the array should be non-negative equal integers.

Your task is to calculate the minimum number of operations you have to perform.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum number of operations you have to perform. If it is impossible to make all elements of the array equal, print -1.

Examples:
input:
3
2
1 1
3
1 3 2
4
2 1 2 6
output:
0
-1
3

Note:

Buggy Code:
import java.io.*;
import java.util.*;
import java.lang.*;
import java.util.function.LongConsumer;

public class CP
{
    static long mod = (long)(1e9+7);
    static int minInt = Integer.MIN_VALUE;
    static int maxInt = Integer.MAX_VALUE;
    static long minLong = Long.MIN_VALUE;
    static long maxLong = Long.MAX_VALUE;

    /*-----------------------------------------------------------------------MAIN CODE HERE----------------------------------------------------------------------------*/
    static void time_is_running(PrintWriter out,FastReader sc) throws IOException
    {
        int n = sc.ni(),arr[] = new int[n];
        for(int j=0;j<n;j++) arr[j] = sc.ni();

        int s=0,e=(int)1e9,ans=maxInt;
        while(s<=e){
            int mid = s+((e-s)/2);

            int tmp = check(arr,mid);
//            out.println(mid+" "+tmp);
            if(tmp==-1){
                e=mid-1;
            }
            else{
                ans=Math.min(ans,tmp);
                s=mid+1;
            }
        }
        ans=ans==maxInt?-1:ans;

        out.println(ans);
    }

    static int check(int a[],int k){
        int maxInd=0;
        long sum=0;
        int arr[] = Arrays.copyOf(a,a.length);
        for(int j=0;j<arr.length;j++){
            if(arr[maxInd]<arr[j]) maxInd=j;
            sum+=arr[j];
        }

        if(arr[maxInd]<k) return -1;
        int totalOps=0,ind=maxInd;
        while(sum>(k*1l*a.length)){
            if(arr[ind]<=k){
                totalOps+=(k-arr[ind]);
                sum-=(k-arr[ind]);
                arr[(ind-1+arr.length)%arr.length]-=(2*(k-arr[ind]));
                arr[ind]=k;
            }
            else{
                int ops = (arr[ind]-k+1)/2;
                sum-=ops;
                totalOps+=ops;
                arr[ind]-=(2*ops);
                arr[(ind+1)%arr.length]+=ops;
            }
            ind=(ind+1)%arr.length;
        }

        for(int e:arr){
            if(e!=k) return -1;
        }

        return totalOps;
    }
    /*----------------------------------------------------------------------------END---------------------------------------------------------------------------------*/

    public static void main(String[] args) throws IOException
    {
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        FastReader sc = new FastReader();
        int n = sc.ni();
        for(int i=1;i<=n;i++)
            time_is_running(out,sc);
        out.close();
    }

    /*-----------------------------------------------------------------------SKIP THIS PART----------------------------------------------------------------------------*/
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
//        public FastReader() throws FileNotFoundException {
//            br = new BufferedReader(new FileReader("input.txt"));
//        }
        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int ni() {return Integer.parseInt(next());}
        long nl() {return Long.parseLong(next());}
        double nd() {return Double.parseDouble(next());}
        char nextChar(){return next().charAt(0);}
        String nextLine() {
            String str = "";
            try {
                if (st.hasMoreTokens()) {
                    str = st.nextToken("\n");
                } else {
                    str = br.readLine();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    /*-------------------------------------------------------------------------END------------------------------------------------------------------------------*/
}
class Pair
{
    int a,b,c;
    Pair(int a1,int b1,int c1)
    {
        this.a=a1;
        this.b=b1;
        this.c=c1;
    }
}
class Pair2
{
    long a,b;
    Pair2(long a1,long b1)
    {
        this.a=a1;
        this.b=b1;
    }
}
class comparingLong implements Comparator<Pair2>{
    public int compare(Pair2 s1, Pair2 s2){
        // We are returning the object in asending order
        if (s1.a < s2.a)
            return -1;
        else if (s1.a > s2.a)
            return 1;
        return 0;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to consider:

1. **Input Size and Constraints**:
   - The array size \( n \) should be within the range \( 2 \le n \le 2 \cdot 10^5 \).
   - The elements of the array \( a_i \) should be within the range \( 1 \le a_i \le 10^9 \).

2. **Edge Cases**:
   - **Equal Elements**: An array where all elements are already equal and non-negative.
   - **Almost Equal Elements**: An array where elements are very close to being equal but require a minimal number of operations.
   - **Large Discrepancies**: An array with large differences between elements.
   - **Sum Constraints**: An array where the sum of elements is not divisible by \( n \), making it impossible to equalize the elements.

3. **Challenging Scenarios**:
   - **Small Array with Large Values**: This can test the limits of the algorithm's ability to handle large values.
   - **Large Array with Small Values**: This can test the efficiency of the algorithm in handling large arrays.
   - **Special Patterns**: Arrays with specific patterns that might cause the algorithm to fail, such as alternating high and low values.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
3
3
1 2 3
4
1 1 1 1
5
10 20 30 40 50
```

### Explanation:
1. **Test Case 1**:
   - Array: [1, 2, 3]
   - This array has elements that are close but not equal. It will test the algorithm's ability to equalize the elements with minimal operations.

2. **Test Case 2**:
   - Array: [1, 1, 1, 1]
   - This array is already equal. It will test the algorithm's ability to handle an already equal array and return 0 operations.

3. **Test Case 3**:
   - Array: [10, 20, 30, 40, 50]
   - This array has large differences between elements. It will test the algorithm's ability to handle large discrepancies and calculate the correct number of operations.

This test input should expose any bugs in the given code, particularly in handling different scenarios and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
#include<cstdio>
using namespace std;
int a[500005],T,n,ans[500005],st[500005],lst;
inline int Mx(int x,int y){return x>y?x:y;}
inline int Mn(int x,int y){return x<y?x:y;}
int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            ans[i]=a[i]<ans[i-1]?ans[i-1]:a[i];
        }
        // int cnt=0;
        lst=0;
        for(int i=n;i;i--){
            st[i]=a[i];
            while(ans[i]==ans[i-1]){
                ans[i-1]=ans[i];
                st[i-1]=Mn(st[i],a[i-1]);
                i--;
            }
            if(a[i]>st[lst])for(int j=i;j<lst;j++)ans[j]=ans[lst];
            lst=i;
        }
        for(int i=1;i<=n;i++)printf("%d ",ans[i]);printf("\n");
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
4
1 2 3 4
4
4 3 2 1
5
1 3 2 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int N=500010;
int n;
int a[N];
int rpos[N];
int mx[N],mxpos[N];
int ans[N];
int tr[1000010];
int f[N];

inline int lowbit(int x)
{
	return x&(-x);
}

int query(int x)
{
	int mx=0;
	while (x>0)
	{
		mx=max(mx,tr[x]);
		x-=lowbit(x);
	}
	return mx;
}

void modify(int x,int v)
{
	while (x<=n)
	{
		tr[x]=max(tr[x],v);
		x+=lowbit(x);
	}
}

int main()
{
	int T;
	cin >> T;
	while (T--)
	{
		cin >> n;
		for (int i=1;i<=n;i++)
		{
			cin >> a[i];
		}

		for (int i=1;i<=n;i++)
		{
			if (a[i]>mx[i-1])
			{
				mx[i]=a[i];
				mxpos[i]=i;
			}
			else
			{
				mx[i]=mx[i-1];
				mxpos[i]=mxpos[i-1];
			}
		}

		memset(tr,0,sizeof(tr));
		for (int i=n;i>=1;i--)
		{
			rpos[i]=max(i,query(a[i]-1));
			modify(a[i],i);
		}

		//memset(f,-1,sizeof(f));
		for (int i=1;i<=n;i++)
		{
			int j=i;
			int mx=a[i];
			while (mxpos[rpos[j]]!=j)
			{
				j=mxpos[rpos[j]];
				mx=max(mx,a[j]);
			}
			cout << mx << " ";
		}
		cout << "\n";
	}
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that might cause the algorithm to fail. Here are some steps to think through this:

1. **Edge Cases**:
   - **Single Tree**: Test cases with only one tree.
   - **All Trees of the Same Height**: Test cases where all trees have the same height.
   - **Increasing Heights**: Test cases where the heights are strictly increasing.
   - **Decreasing Heights**: Test cases where the heights are strictly decreasing.
   - **Mixed Heights**: Test cases with a mix of increasing and decreasing heights.

2. **Critical Scenarios**:
   - **Backward Jumps**: Scenarios where the rabbit can only jump backward.
   - **Forward Jumps**: Scenarios where the rabbit can only jump forward.
   - **Complex Jumps**: Scenarios where the rabbit needs to make a series of jumps, both forward and backward, to reach the maximum height.

3. **Input Format**:
   - The input should follow the format specified in the problem description.

Let's create a test input that combines some of these edge cases and critical scenarios:

```plaintext
3
1
5
5
5 4 3 2 1
4
1 3 2 4
```

### Explanation:
1. **Single Tree**:
   - This test case checks if the algorithm handles a single tree correctly.
   - Input: `1 5`
   - Expected Output: `5`

2. **Decreasing Heights**:
   - This test case checks if the algorithm can handle a strictly decreasing sequence of tree heights.
   - Input: `5 5 4 3 2 1`
   - Expected Output: `5 5 5 5 5`

3. **Mixed Heights**:
   - This test case checks if the algorithm can handle a mix of increasing and decreasing heights.
   - Input: `4 1 3 2 4`
   - Expected Output: `4 4 4 4`

This input is small-scale and complete, and it should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
#include<iostream>
using namespace std;
int main()
{
	int n;
	cin>>n;
	for (int p=0;p<n;p++)
	{

		if(p==0)cout<<3<<" "<<3<<" "<<3<<" "<<4<<endl;
		if(p==1)cout<<5<<" "<<5<<" "<<5<<" "<<5<<" "<<5<<endl;
		if(p==2)cout<<2<<" "<<2<<" "<<2<<" "<<3<<endl;
		if(p==3)cout<<1<<" "<<1<<" "<<3<<" "<<3<<endl;
		if(p==4)cout<<8<<" "<<8<<" "<<8<<" "<<8<<" "<<8<<" "<<8<<" "<<8<<" "<<8<<endl;

	}
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
4
2 3 1 4
5
5 4 3 2 1
8
2 4 1 6 3 8 5 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import random

def adjust(list_n,xiabiao,flag):
    global list_max,index_list
    max_tmp=list_n[xiabiao]
    #print(list_max)
    list_max.append(max_tmp)
    for j in range(len(list_n)):
        if xiabiao>j and list_n[xiabiao]<list_n[j] and flag==1:
            list_max.append(list_n[j])
            index_list.append(j)
            adjust(list_n, j, 1)
        if xiabiao<j and list_n[xiabiao]>list_n[j] and flag ==0:
            index_list.append(j)
            adjust(list_n, j, 0)
    return

if __name__ == '__main__':
    num=int(input())
    for i in range(num):
        # 获取总个数
        n=int(input())
        # 获取数组
        user_input = input()
        number_strings = user_input.split()
        list_n = [int(n0) for n0 in number_strings]
        #print(list_n)
        for w in range(len(list_n)) :
            list_max=[list_n[w]]
            index_list=[w]
            for k in range(len(list_n)):
                adjust(list_n, w, 0)
            for k in range(len(list_n)):
                adjust(list_n, max(index_list), 1)

            for ww in range(len(list_n)*5):
                random1= random.randint(min(index_list), max(index_list))
                random2 = random.randint(min(index_list), max(index_list))
                for k in range(len(list_n)) :
                   adjust(list_n,random1,0)
                for k in range(len(list_n)) :
                    adjust(list_n,random2, 1)
            print(max(list_max),end=' ')
        print()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that can cause the algorithm to fail. The key issues in the buggy code are:

1. **Recursive Adjustment**: The `adjust` function is called recursively, which can lead to deep recursion or incorrect traversal.
2. **Randomization**: The code uses random indices to adjust the list, which can lead to non-deterministic results and potentially miss optimal solutions.
3. **Flag Management**: The `flag` parameter is used to determine the direction of the jump, but the logic is not robust enough to handle all cases.

### Step-by-Step Analysis

1. **Edge Case 1: All Trees of the Same Height**
   - If all trees have the same height, the rabbit should not be able to jump anywhere. This can expose issues with the recursive calls and randomization.

2. **Edge Case 2: Trees in Ascending Order**
   - If the trees are in ascending order, the rabbit can only jump backward to taller trees. This can test the backward jump logic.

3. **Edge Case 3: Trees in Descending Order**
   - If the trees are in descending order, the rabbit can only jump forward to shorter trees. This can test the forward jump logic.

4. **Edge Case 4: Alternating Heights**
   - If the heights alternate between two values, the rabbit can jump back and forth between the two heights. This can test the recursive logic and ensure it handles such patterns correctly.

5. **Edge Case 5: Single Tree**
   - If there is only one tree, the rabbit should not be able to jump anywhere. This can test the base case handling.

### Chosen Test Input

Given the constraints and the need for a small-scale test input, I will create a test case with 5 trees that includes a mix of the above edge cases:

- **Test Case 1**: All trees of the same height.
- **Test Case 2**: Trees in ascending order.
- **Test Case 3**: Trees in descending order.
- **Test Case 4**: Alternating heights.
- **Test Case 5**: Single tree.

### Output

```plaintext
5
5
1 1 1 1 1
5
1 2 3 4 5
5
5 4 3 2 1
5
1 2 1 2 1
1
3
```

This test input will help identify issues with the recursive calls, randomization, and flag management in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
def solve_rabbit_jumps():
    t = int(input())

    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))

        def dfs(start, visited):
            max_height = a[start]
            visited[start] = True

            # 向左跳（跳到更高的树）
            for j in range(start-1, -1, -1):
                if not visited[j] and a[j] > a[start]:
                    max_height = max(max_height, dfs(j, visited))

            # 向右跳（跳到更矮的树）
            for j in range(start+1, n):
                if not visited[j] and a[j] < a[start]:
                    max_height = max(max_height, dfs(j, visited))

            return max_height

        # 对每个起始位置计算最大高度
        result = []
        for i in range(n):
            visited = [False] * n
            max_height = dfs(i, visited)
            result.append(max_height)

        print(*result)

# 运行程序
solve_rabbit_jumps()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
4
1 2 3 4
4
4 3 2 1
5
1 3 2 4 5
```

### Explanation:
1. **Test Case 1:**
   - Input: `4 1 2 3 4`
   - Output: `4 4 4 4`
   - **Reason:** The rabbit can always jump to the right to a taller tree. The maximum height is 4 for all starting positions.

2. **Test Case 2:**
   - Input: `4 4 3 2 1`
   - Output: `4 4 4 4`
   - **Reason:** The rabbit can always jump to the left to a taller tree. The maximum height is 4 for all starting positions.

3. **Test Case 3:**
   - Input: `5 1 3 2 4 5`
   - Output: `5 5 5 5 5`
   - **Reason:** The rabbit can jump to the fifth tree from any starting position. The maximum height is 5 for all starting positions.

These test cases cover scenarios where the heights are strictly increasing, strictly decreasing, and mixed, which can help expose any issues in the DFS logic and ensure the algorithm handles all edge cases correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import random

def adjust(list_n,xiabiao,flag):
    global list_max,index_list
    max_tmp=list_n[xiabiao]
    #print(list_max)
    list_max.append(max_tmp)
    for j in range(len(list_n)):
        if xiabiao>j and list_n[xiabiao]<list_n[j] and flag==1:
            list_max.append(list_n[j])
            index_list.append(j)
            adjust(list_n, j, 1)
        if xiabiao<j and list_n[xiabiao]>list_n[j] and flag ==0:
            index_list.append(j)
            adjust(list_n, j, 0)
    return

if __name__ == '__main__':
    num=int(input())
    for i in range(num):
        # 获取总个数
        n=int(input())
        # 获取数组
        user_input = input()
        number_strings = user_input.split()
        list_n = [int(n0) for n0 in number_strings]
        #print(list_n)
        for w in range(len(list_n)) :
            list_max=[list_n[w]]
            index_list=[w]
            for k in range(len(list_n)):
                adjust(list_n, w, 0)
            for k in range(len(list_n)):
                adjust(list_n, max(index_list), 1)

            for ww in range(len(list_n)*10):
                random1= random.randint(min(index_list), max(index_list))
                random2 = random.randint(min(index_list), max(index_list))
                for k in range(len(list_n)) :
                   adjust(list_n,random1,0)
                for k in range(len(list_n)) :
                    adjust(list_n,random2, 1)
            print(max(list_max),end=' ')
        print()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
5
1 2 3 4 5
5
5 4 3 2 1
8
4 3 2 1 5 6 7 8
```

### Explanation:
1. **Test Case 1: Strictly Increasing Sequence**
   - **Input:** `1 2 3 4 5`
   - **Output:** `5 5 5 5 5`
   - **Explanation:** In this case, the rabbit can always jump forward to the next taller tree, eventually reaching the tallest tree.

2. **Test Case 2: Strictly Decreasing Sequence**
   - **Input:** `5 4 3 2 1`
   - **Output:** `5 5 5 5 5`
   - **Explanation:** In this case, the rabbit can always jump backward to the next taller tree, eventually reaching the tallest tree.

3. **Test Case 3: Mixed Sequence with a Peak**
   - **Input:** `4 3 2 1 5 6 7 8`
   - **Output:** `8 8 8 8 8 8 8 8`
   - **Explanation:** The sequence has a decreasing part followed by an increasing part. The rabbit can jump backward from the first four trees to the last four trees, and from the last four trees, it can jump forward to the tallest tree.

These test cases are designed to cover different scenarios:
- A strictly increasing sequence to test the forward jumping logic.
- A strictly decreasing sequence to test the backward jumping logic.
- A mixed sequence to test both forward and backward jumping logic, and to ensure the algorithm can handle a peak in the sequence.

By using these test cases, we can effectively expose any bugs in the buggy code, especially in the recursive adjustment function.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import java.util.*;
import java.io.*;

public class codeforces {

    static int MOD =  998244353 ;
    static long modInverse(long x, int MOD) {
        return power(x, MOD - 2, MOD);
    }

    static long power(long base, long exp, int MOD) {
        long result = 1;
        while (exp > 0) {
            if ((exp & 1) == 1) { // If exp is odd
                result = (result * base) % MOD;
            }
            base = (base * base) % MOD;
            exp >>= 1; // Divide exp by 2
        }
        return result;
    }

    static int max = 0 ;
    static void dfs(int i, int p,int[] a ,int[] visited, int[] dp, int n) {
        visited[i] = 1 ;

        max = Math.max(max, a[i]) ;

        for (int j = 0; j < i; j++) {
            if(a[j] > a[i] && visited[j] == 0) dfs(j,i,a,visited,dp,n) ;
        }
        for (int j = i+1; j < n; j++) {
            if(a[j] < a[i] && visited[j] == 0)  dfs(j,i,a,visited,dp,n) ;
        }

        dp[i] = max ;
    }

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt(); // Number of test cases
        while (t-- > 0) {
            int n = sc.nextInt() ;
            max = 0 ;
            int[] a = new int[n]  ;
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt() ;
            }

            int[] ans = new int[n] , v = new int[n] ;
            for (int i = 0; i < n; i++) {
                if(v[i] == 0) dfs(i,-1,a,v,ans,n) ;
            }
            for(int x : ans) System.out.print(x+" ");
            System.out.println();
        }
        sc.close();
    }

}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
4
1 2 3 4
5
5 4 3 2 1
6
1 3 2 4 5 6
```

**Explanation:**

1. **Test Case 1:**
   - **Input:** `4\n1 2 3 4`
   - **Expected Output:** `4 4 4 4`
   - **Reason:** The rabbit can jump forward to the next tree if it is taller, but since the sequence is strictly increasing, the rabbit can always jump to the last tree, which has the maximum height of 4.

2. **Test Case 2:**
   - **Input:** `5\n5 4 3 2 1`
   - **Expected Output:** `5 5 5 5 5`
   - **Reason:** The rabbit can jump backward to the first tree, which has the maximum height of 5. This is because the sequence is strictly decreasing.

3. **Test Case 3:**
   - **Input:** `6\n1 3 2 4 5 6`
   - **Expected Output:** `6 6 6 6 6 6`
   - **Reason:** The rabbit can jump forward to the next tree if it is taller, and since the sequence has a strictly increasing subsequence (4, 5, 6), the rabbit can always reach the last tree, which has the maximum height of 6.

These test cases are designed to expose issues in the buggy code, particularly in handling sequences that are strictly increasing, strictly decreasing, and mixed. The buggy code might fail to correctly handle these scenarios, especially the mixed sequence, where the rabbit can jump both forward and backward.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import java.io.*;
import java.util.*;

public class CodeForcesSolution {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static StringTokenizer st;
    static int MOD = (int) 998244353, n;
    static int[] a;
    static List<Integer> lst;
    static int [][] minTable;
    public static void main(String[] args) throws IOException {
        int t = readInt();
        while (t-- > 0) {
            solve();
        }
    }
    static void solve() throws IOException {
        n = readInt();
        a = new int[n]; lst = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            a[i] = readInt();
            lst.add(i);
        }
        minTable = new int[n+1][(int) (Math.log(n)/Math.log(2))+1];
        for (int i = 1; i <= n; i++) {
            minTable[i][0] = a[i - 1];
        }
        for (int i = 1; (1<<i) <= n; i++) {
            for (int j = 1; j+(1<<i)-1 <= n; j++) {
                minTable[j][i] = Math.min(minTable[j][i-1], minTable[j+(1<<(i-1))][i-1]);
            }
        }
        lst.sort((f, s) -> {
            if (a[f] == a[s]) return f - s;
            else return a[s] - a[f];
        });
        List<int[]> cutoff = new ArrayList<>();
        int j = lst.get(0), j2 = n - 1, v = a[lst.get(0)];
        for (int i = 0; i < n; i++) {
            if (j <= lst.get(i) || minQuery(j + 1, j2) < a[lst.get(i)]) {
                j = Math.min(j, lst.get(i));
            } else {
                cutoff.add(new int[] {j, v});
                j2 = j;
                j = lst.get(i);
                v = a[lst.get(i)];
            }
        }
        cutoff.add(new int[] {j, v});
        int i = n - 1;
        int[] ans = new int[n];
        for (int[] c : cutoff) {
            while (i >= c[0]) {
                ans[i--] = c[1];
            }
        }
        printArray(ans, n);
    }
    static int minQuery(int l, int r) {
        if (l > r) return 1<<30;
        int j = (int) (Math.log(r-l+1)/Math.log(2));
        return Math.min(minTable[l][j], minTable[r-(1<<j)+1][j]);
    }
    static void printArray(int[] arr, int size) {
        for(int i = 0; i < size; i++) {
            System.out.print(arr[i]);
            if (i < size - 1) System.out.print(" ");
        }
        System.out.println();
    }
    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
    static long readLong() throws IOException {
        return Long.parseLong(next());
    }
    static int readInt() throws IOException {
        return Integer.parseInt(next());
    }
    static double readDouble() throws IOException {
        return Double.parseDouble(next());
    }
    static char readCharacter() throws IOException {
        return next().charAt(0);
    }
    static String readLine() throws IOException {
        return br.readLine().trim();
    }
    static int readALotInt() throws IOException{
        int x = 0, c;
        while((c = br.read()) != ' ' && c != '\n')
            x = x * 10 + (c - '0');
        return x;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code uses a segment tree (or sparse table) to find the minimum value in a range, which is used to determine the possible jumps of the rabbit. The issue might be in the way the segment tree is built or queried, or in the logic that determines the jumps.

2. **Edge Cases**:
   - **Single Tree**: A single tree should always be the maximum height for that tree.
   - **All Trees of the Same Height**: If all trees have the same height, the rabbit should not be able to jump anywhere.
   - **Monotonically Increasing or Decreasing Heights**: These sequences can test the boundary conditions of the segment tree.
   - **Mixed Heights with Ties**: This can test the sorting and comparison logic.

3. **Challenging Scenarios**:
   - **Mixed Heights with Complex Jump Conditions**: A sequence where the rabbit can jump back and forth multiple times.
   - **Large Range with Small Height Differences**: This can stress the segment tree and the comparison logic.

4. **Generate the Test Input**:
   - We will create a test case with a small number of trees (to keep the input size manageable) but with heights that can expose the bug in the segment tree and the jump logic.

Here is a challenging test input:

```plaintext
3
5
5 4 3 2 1
5
1 2 3 4 5
6
3 1 4 1 5 2
```

Explanation:
- **Test Case 1**: All trees are in decreasing order. The rabbit should be able to jump to the first tree from any starting point.
- **Test Case 2**: All trees are in increasing order. The rabbit should not be able to jump forward.
- **Test Case 3**: A mix of heights that can create complex jump conditions. The rabbit can jump back and forth, and this will test the segment tree and the logic for determining the maximum height.

This test input should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define endl '\n';
#define int ll
#define double long double

vector<int> ans;
vector<vector<int>> gr;
vector<int> used;

void dfs(int i)
{
    used[i] = 1;
    for (auto c : gr[i])
    {
        if (!used[c])
        {
            dfs(c);
        }
        ans[i] = max(ans[i], ans[c]);
    }
}

void solve()
{
    int n;
    cin >> n;
    vector<int> a(n+1);

    for (int i = 1; i <= n; ++i)
    {
        cin >> a[i];
        // a[i]--;
    }

    ans.clear();
    ans.resize(n+1);
    gr.clear();
    gr.resize(n + 2);
    used.clear();
    used.resize(n + 2);
    for (int i = 1; i <= n; ++i)
    {
        ans[i] = a[i];
    }
    vector<pair<int,int>> pref(n + 1);
    vector<int> suf(n + 2, a[n]);
    for (int i = 1; i <= n; ++i)
    {
        if (pref[i-1].first<a[i]){
            pref[i]={a[i],i};
        }
        else{
            pref[i]=pref[i-1];
        }
        // pref[i] = max(pref[i - 1], a[i - 1]);
    }
    for (int i = n; i >= 1; --i)
    {
        suf[i] = min(suf[i + 1], a[i]);
    }
    suf[0] = suf[1];
    //for (auto c : suf)
     //{
      //  cout << c << " ";
    //}

    //cout<<endl;

    for (int i = 1; i <=n; ++i)
    {
        // cout<<a[i]+1<<" "<<suf[i+2]+1<<" "<<pref[i]+1<<endl;
        if (i + 1 < suf.size() && suf[i+1] < a[i])
        {
            // int l=i+1;
            int l = i + 1;
            int r = suf.size();
            while (r - l > 1)
            {
                int mid = (r + l) / 2;
                if (suf[mid] <= a[i])
                {
                    l = mid;
                }
                else
                {
                    r = mid;
                }
            }
            // cout << l << " " << a[i] + 1 << " " << suf[l] + 1 << endl;
            //cout<<l<<" "<<suf[l]<<endl;
            if (suf[l] < a[i] && l<=n)
            {

                //cout<<"suf"<<i+1<<" "<<l<<endl;
                gr[i].push_back(l);

                // gr[suf[l]].push_back(a[i]);
            }
        }
        if (i >= 1 && pref[i].first > a[i])
        {
            // cout<<a[i]+1<<" "<<pref[i]+1<<endl;
            //cout<<"pr"<<i+1<<" "<<pref[i].second+1<<endl;
            gr[i].push_back(pref[i].second);
            // gr[pref[i]].push_back(a[i]);
        }

        dfs(i);
    }
    for (int k=0;k<n;++k){
        for (int i=1;i<=n;++i){
            dfs(i);
        }
    }

    // for (auto c:gr[5]){
    //     cout<<c+1<<"w";
    // }
    // cout<<endl;
    // for (int i = 0; i < n; ++i)
    // {
    //     cout << i + 1 << " " << ans[i] + 1 << endl;
    // }
    // for ()

    for (int i = 1; i <= n; ++i)
    {
        cout << ans[i] << " ";
    }
    cout << endl;
}

signed main()
{
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5
5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
#pragma GCC optimize("O3,unroll-loops")
#include<bits/stdc++.h>
// #include <atcoder/all>
// #include<ext/pb_ds/assoc_container.hpp>
// #include<ext/pb_ds/tree_policy.hpp>
using namespace std;

// using namespace __gnu_pbds;
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

// https://docs.google.com/document/d/1VuLkJfkaBBFdcV-BPtT2JdTDrD2lC3POFj6NPTl_Pzo/edit?addon_store&tab=t.0
// typedef tree<int, null_type, greater<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; // find_by_order, order_of_key
//find by order-> element at ith idx(starting from 0)//gives iterator to that index//if not present *it gives 0
//order of key-> number of element less than a number

// erase function for ordered multiset
// void myerase(pbds &t, int v){
//     int rank = t.order_of_key(v);//Number of elements that are less than v in t
//     pbds::iterator it = t.find_by_order(rank); //Iterator that points to the (rank+1)th element in t
//     t.erase(it);
// }

#define int long long int
// #define ll long long int
#define loop(var, s, e) for(int var = s; var < e; var++)
#define rloop(var, e, s) for(int var = e; var>= s; var--)
#define v vector<int>
// #define vll vector<long long int>
#define printyes cout<<"YES\n"
#define printno cout<<"NO\n"
//COMMENT THIS FOR INTERACTIVE PROBLEMS...............
// #define endl '\n'
#define mod (int)(998244353)
#define inf 1e18
#define dbg(x) cerr<<"value of "<<#x<<" is => "<<x<<endl;
#define all(x) x.begin(), x.end()
#define vpii vector<pair<int, int>>
#define pii pair<int,int>
#define pll pair<ll, ll>
#define ff first
#define ss second
#define pb push_back
#define vpll vector<pll>
#define here(i) cerr<<"here "<<i<<" "<<endl;
#define rt return;
#define mx max_element
#define mn min_element
#define asum accumulate
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());
template <class A, class B> ostream& operator << (ostream& out, const pair<A, B> &a) {
	return out << "(" << a.ff << ", " << a.ss << ")";
}

template <class A> ostream& operator << (ostream& out, const vector<A> &ve) {
	out << "[";
	loop(i, 0, ve.size()) {
		if(i) out << ", ";
		out << ve[i];
	}
	return out << "]";
}

                //U R D L
int delrow[4] = {-1,0,1,0};
int delcol[4] = {0,1,0,-1};
// unordered_map<char, int> mpp = {{'U', 0}, {'R', 1}, {'D', 2}, {'L', 3}};

namespace bit {
  int setBit(int n, int bitNo) {
    n = (n|(1LL<<bitNo));
    return n;
  }

  int currBit(int n, int bitNo) {
    return ((n&(1ll<<bitNo)) == 0)?0:1;
  }

  int unSetBit(int n, int bitNo) {
    n = (n&(~(1ll<<bitNo)));
    return n;
  }
  /*
  __lg() returns the highest set bit index in a number i.e for 100 it will be 2 and for 000 will be -1

  n&(n - 1) clears the first set bit

  n&(-n) give the first set bit => 101100 -> 000100

  n&(n + 1) clear all trailing ones

  n|(n + 1) set the first unset bit
  */
}
//*************************************************CHECK THE CONSTRAINTS**********************************************************//
int extendedEuclidean(int a, int b, int&x, int& y) {
  if(b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int x1, y1;
  int d = extendedEuclidean(b, a%b, x1, y1);
  x = y1;
  y = x1 - y1*(a/b);
  return d;
}

int f(int a, int b) {
  if(b == 0)return 1;
  int res = f(a, b/2);
  if(b%2 != 0) {
    return (((res)%mod* (res)%mod)%mod * a)%mod;
  }
  else {
    return ((res)%mod * (res)%mod)%mod;
  }
}

int _f(int a, int b) {
  if(b == 0)return 1;
  int res = f(a, b/2);
  if(b%2 != 0) {
    return (((res)* (res))* a);
  }
  else {
    return ((res) * (res));
  }
}

int inv(int a, int b){
 return 1<a ? b - inv(b%a,a)*b/a : 1;
}

vector<bool> sieve(int n) {
  vector<bool> is_prime(n+1, 1);
  // v prime;
  // v dp(n + 1, 0);
  is_prime[0] = 0;
  is_prime[1] = 0;
  for(int i = 2; i <= n; i++) {
    if(is_prime[i] == 1) {
      // prime.pb(i);
      for(int j = i*i; j <= n; j+=i) {
        is_prime[j] = 0;
      }
    }
  }
  return is_prime;
}
// void *memcpy(void *dest, const void *src, size_t n);
int ceilDiv(int a, int b) {
  double _a = a;
  int ans = ceil(_a/b);
  return ans;
}

// first write complete pseudo code on paper then code dont care about the time!!!!!!

void pikachu(int tt, int tc) {
  int n;
  cin>>n;
  v arr(n);
  loop(i, 0, n)cin>>arr[i];
  int min_so_far = inf;
  v lastMinIndex(n + 1, inf);
  rloop(i, n - 1, 0) {
    if(arr[i] < min_so_far) {
      min_so_far = arr[i];
      loop(j, arr[i] + 1, n + 1) {
        if(lastMinIndex[j] != inf)break;
        lastMinIndex[j] = i;
      }
    }
  }

  v ans(n, inf);
  v prevMaxIndex(n);
  prevMaxIndex[0] = 0;
  loop(i, 1, n) {
    if(arr[i] <= arr[prevMaxIndex[i - 1]])prevMaxIndex[i] = prevMaxIndex[i - 1];
    else prevMaxIndex[i] = i;
  }
  auto dfs = [&](int src, auto& self)->int {
    if(ans[src] != inf)return ans[src];
    ans[src] = arr[src];
    if(lastMinIndex[arr[src]] != inf)ans[src] = max(ans[src], self(lastMinIndex[arr[src]], self));
    if(prevMaxIndex[src] != src)ans[src] = max(ans[src], self(prevMaxIndex[src], self));
    return ans[src];
  };
  loop(i, 0, n)dfs(i, dfs);
  for(auto it: ans)cout<<it<<" ";
  cout<<endl;
}

int32_t main() {
  auto begin = std::chrono::high_resolution_clock::now();
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  #ifndef ONLINE_JUDGE
      freopen("input.txt", "r", stdin);
      freopen("output.txt", "w", stdout);
  #endif
  int gsBall = 1;
  cin>>gsBall;
  int tt = gsBall;
  int tc = 1;
  cout<<setprecision(20);
  while(gsBall--) {
    pikachu(tt, tc);
    tc++;
  }
  auto end = std::chrono::high_resolution_clock::now();
  auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
  cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds.\n";
  return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes potential bugs in the given code, we need to focus on edge cases and scenarios that could cause issues. Here are some considerations:

1. **Single Tree**: A test case with only one tree should always return the height of that tree.
2. **All Trees of the Same Height**: A test case where all trees have the same height should return the same height for each tree.
3. **Monotonically Increasing or Decreasing Heights**: A test case with heights that are strictly increasing or decreasing should be handled correctly.
4. **Mixed Heights**: A test case with a mix of increasing and decreasing heights to test the logic for both forward and backward jumps.
5. **Large Heights**: Test cases with large values to ensure the algorithm handles large inputs correctly.

Given these considerations, let's create a small-scale test input that includes a mix of these scenarios:

```plaintext
3
1
5
4
5 4 3 2 1
6
1 3 2 5 4 6
```

### Explanation:
1. **Test Case 1**: A single tree with height 5.
   - Expected output: `5`

2. **Test Case 2**: Four trees with heights in strictly decreasing order.
   - Expected output: `5 5 5 5`

3. **Test Case 3**: Six trees with mixed heights.
   - Expected output: `6 6 6 6 6 6`

This test input covers a variety of scenarios and should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import math
from sys import stdout

def ii():
    return int(input())

def mi():
    return map(int, input().split())

def li():
    return list(mi())

class UF:

    def __init__(self, n):
        self.p = [i for i in range(n)]
        self.ranks = [0] * n

    def find(self, x):
        if x == self.p[x]: return x
        self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, x, y):
        x, y = self.find(x), self.find(y)
        if x == y: return
        if self.ranks[x] > self.ranks[y]: self.p[y] = x
        elif self.ranks[x] < self.ranks[y]: self.p[x] = y
        else: self.p[y] = x; self.ranks[x] += 1

def bin_right(arr, curr, uf):
    l = 0; r = len(arr) - 1
    while l <= r:
        mid = (l + r) // 2
        if arr[mid][0] < curr[0]:
            if arr[mid][1] > curr[1]:
                uf.union(curr[1], arr[mid][1])
        r = mid - 1

    return uf

def bin_left(arr, curr, uf):
    l = 0; r = len(arr) - 1
    while l <= r:
        mid = (l + r) // 2
        if arr[mid][0] > curr[0]:
            if arr[mid][1] < curr[1]: uf.union(curr[1], arr[mid][1])
        l = mid + 1
    return uf

def solve():
    n = ii(); a = li(); uf = UF(n)
    array = [(a[i], i) for i in range(n)]; array.sort()

    for i in range(n):
        uf = bin_left(array, (a[i], i), uf)
        uf = bin_right(array, (a[i], i), uf)

    segments = []; i = 0; j = 0
    while j < n:
        while j < n and uf.p[i] == uf.p[j]: j += 1
        segments.append((i, j - 1)); i = j

    ans = list()
    for s in segments:
        curr = max(a[s[0]: s[1] + 1])
        ans += [f'{curr}'] * (s[1] - s[0] + 1)
    print(' '.join(ans))

t = int(input())
for _ in range(t):
    solve()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
4
4 3 2 1
5
1 2 3 4 5
6
3 1 4 1 5 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import sys
input = sys.stdin.readline
from collections import deque
from bisect import bisect_left

def dfs(i):
    if dp[i] != -1: return dp[i]

    dp[i] = li[i]
    dp[i] = max(dp[i], dfs(b[i]))
    dp[i] = max(dp[i], dfs(f[i]))

    return dp[i]

def solve():
    global n, b, f, dp, li

    n = int(input())
    li = list(map(int, input().split()))
    a = [(li[i], i) for i in range(n)]
    dp = [-1]*n
    ans = [0]*n
    f = [0]*n
    b = [0]*n

    st = deque()
    st.appendleft((int(1e9), -1))
    for i in range(n - 1, -1, -1):
        if li[i] < st[0][0]:
            st.appendleft((li[i], i))

        j = bisect_left(st, (li[i], 1e9)) - 1
        if li[i] == st[j][0]:
            f[i] = i
        else:
            f[i] = st[j][1]

    # print(f)

    a.sort(reverse=True)
    r = n
    for i in range(n):
        for j in range(a[i][1], r):
            b[j] = a[i][1]
        r = min(r, a[i][1])

    # print(*f)
    # print(*b)

    for i in range(n):
        ans[i] = dfs(i)

    print(*ans)

for _ in range(int(input())):
    solve()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
4
1 2 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import java.util.Scanner;

public class Main {
    static int maxHeight(int index, int n, int[] heights, int[] memo, boolean[] visited) {
        if (visited[index]) {
            return memo[index];
        }
        visited[index] =true;
        int max = heights[index];
        max = Math.max(max, backJumps(index, n, heights, memo, visited));
        max = Math.max(max, forwardJumps(index, n, heights, memo, visited));

        memo[index] = max;
        visited[index] = false;
        return max;
    }

    static int backJumps(int index, int n, int[] heights, int[] memo, boolean[] visited) {
        int max = heights[index];
        for (int i = index-1; i > -1; i--) {
            if (heights[i] > heights[index]) {
                int height = maxHeight(i, n, heights, memo, visited);
                max = Math.max(max, height);
            }
        }
        return max;
    }

    static int forwardJumps(int index, int n, int[] heights, int[] memo, boolean[] visited) {
        int max = heights[index];
        for (int i = index + 1; i < n; i++) {
            if (heights[i] < heights[index]) {
                int height = maxHeight(i, n, heights, memo, visited);
                max = Math.max(max, height);
            }
        }
        return max;
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        int t = scan.nextInt();

        for (int test = 0; test < t; test++) {
            int n = scan.nextInt();
            int[] heights = new int[n];
            int[] memo = new int[n];
            boolean[] visited = new boolean[n];

            for (int i = 0; i < n; i++) {
                heights[i] = scan.nextInt();
                memo[i] = -1;
                visited[i] = false;
            }

            for (int i = 0; i<n; i++) {
                System.out.print(maxHeight(i, n, heights, memo, visited) + " ");
            }
            System.out.println();
        }

    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
5
5 4 3 2 1
5
1 2 3 4 5
5
3 1 4 2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Input: The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$.

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import java.util.Scanner;

public class Forked {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();  // Number of test cases
        while (t-- > 0) {
            int n = sc.nextInt();  // Length of permutation
            int[] p = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = sc.nextInt();
            }

            for(int i = 0; i < n ; i++){
                int currval = p[i];
                int max = currval;
                int minidx = -1;

                for(int j = i + 1 ; j < n ; j++){
                    max = Math.max(max, p[j]);
                    if(p[j] < currval){
                        minidx = j;
                        currval = max;
                    }
                }
                if(minidx!=-1){
                    for(int k = i ; k<=minidx; k++){
                        p[k] = currval;
                    }
                    i = minidx;
                }
                else{
                    break;
                }
            }
            int maxidx = n-1;
            int max = p[n-1];
            for(int i = n-2 ; i >= 0 ; i--){
                if(p[i] > max){
                    max = p[i];
                    maxidx = i;
                }
            }
            for(int i = n-1 ; i>= maxidx ; i--){
                p[i] = max;
            }
            for(int i = 0; i < n ; i++){
                System.out.print(p[i] + " ");
            }
            System.out.println();
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
5
1 2 3 4 5
5
5 4 3 2 1
5
3 1 4 2 5
```