**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (easy version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $$$r=l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$. Note that in this version, you are only required to output $$$f([a_l, a_{l+1}, \ldots, a_{l+k-1}])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r=l+k-1$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the first query of the first testcase, $$$b=[1,2,3,2,1]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$2$$$ moves:

- Set $$$b_4=4$$$
- Set $$$b_5=5$$$

In the second query of the first testcase, $$$b=[2,3,2,1,2]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$3$$$ moves:

- Set $$$b_3=0$$$
- Set $$$b_2=-1$$$
- Set $$$b_1=-2$$$

Buggy Code:
def get_longest_consecutive_subarray(arr):
    if len(arr) == 0:
        return 0
    longest_length = 1
    current_length = 1
    for i in range(1, len(arr)):
        if arr[i] - arr[i - 1] == 1:
            current_length += 1
        else:
            longest_length = max(longest_length, current_length)
            current_length = 1
    longest_length = max(longest_length, current_length)
    return longest_length

def solve():
    t = int(input())
    for _ in range(t):
        n, k, q = map(int, input().split())
        a = list(map(int, input().split()))
        for _ in range(q):
            l, r = map(int, input().split())
            l -= 1
            r -= 1
            subarray = a[l:r + 1]
            longest_consecutive_length = get_longest_consecutive_subarray(subarray)
            operations_required = max(0, k - longest_consecutive_length)
            print(operations_required)

solve()

# 0kT81M2KE3d4e9xfsco

**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
2
5 3 2
1 2 3 1 2
1 3
2 4
7 4 3
4 5 6 7 8 9 10
1 4
2 5
3 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (easy version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $$$r=l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$. Note that in this version, you are only required to output $$$f([a_l, a_{l+1}, \ldots, a_{l+k-1}])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r=l+k-1$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the first query of the first testcase, $$$b=[1,2,3,2,1]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$2$$$ moves:

- Set $$$b_4=4$$$
- Set $$$b_5=5$$$

In the second query of the first testcase, $$$b=[2,3,2,1,2]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$3$$$ moves:

- Set $$$b_3=0$$$
- Set $$$b_2=-1$$$
- Set $$$b_1=-2$$$

Buggy Code:
/*
ziad-abdou1

    ┌──────────────────────────────────────────────────┐
    │                                                  │
    │      ________                   __               │
    │     /\_____  \  __             /\ \              │
    │     \/____//'/'/\_\     __     \_\ \             │
    │          //'/' \/\ \  /'__`\   /'_` \            │
    │         //'/'___\ \ \/\ \L\.\_/\ \L\ \           │
    │         /\_______\ \_\ \__/.\_\ \___,_\          │
    │         \/_______/\/_/\/__/\/_/\/__,_ /          │
    │                                                  │
    └──────────────────────────────────────────────────┘

*/

import java.io.*;
import java.lang.*;
import java.util.*;

public class Main {
    public static Scanner sc = new Scanner(System.in);
    public static PrintWriter pw = new PrintWriter(System.out);

    static int[] dx = {1, 0, -1, 0};
    static int[] dy = {0, 1, 0, -1};
    //    static arrList<Integer>[] g;
    static int[] larg;
    static int mod = (int) 1e9 + 7;

    public static void main(String[] args) throws IOException {

        int test = 1;

        test = sc.nextInt();
        while (test-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            int q = sc.nextInt();
            long[] a = new long[n];
            long[] ans = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextLong() - i;
            }

            HashMap<Long, Integer> freq = new HashMap<>();
            TreeSet<Pair> freq2 = new TreeSet<>();

            for (int i = 0; i < k - 1; i++) {
                long val = freq.getOrDefault(a[i], 0);
                freq.put(a[i], (int) (val + 1));
                freq2.remove(new Pair(val, a[i]));
                freq2.add(new Pair(val + 1, a[i]));
            }

            for (int i = k - 1; i < n; i++) {
                long val = freq.getOrDefault(a[i], 0);
                freq.put(a[i], (int) (val + 1));
                freq2.remove(new Pair(val, a[i]));
                freq2.add(new Pair(val + 1, a[i]));
                ans[i] = k - freq2.last().a;
                val = freq.getOrDefault(a[i - k + 1], 0);
                freq.put(a[i - k + 1], (int) (val - 1));
                freq2.remove(new Pair(val, a[i - k + 1]));
                freq2.add(new Pair(val - 1, a[i - k + 1]));
            }

            while (q-- > 0) {
                int l = sc.nextInt();// ملهوش لازمة
                int r = sc.nextInt() - 1;
                pw.println(ans[r]);
            }

        }

        pw.flush();

    }

    static long fastpow(long n, long ti) {
        if (ti == 0) return 1l;
        if (ti % 2 == 0) {
            long y = fastpow(n, ti / 2);
            long k = y * y;
            k %= mod;
            return k;
        } else {
            long y = fastpow(n, ti / 2);
            long k = ((n * y) % mod) * y;
            k %= mod;
            return k;
        }
    }

    static long modInverse(long x) {
        return fastpow(x, mod - 2);
    }

    static int lcm(int a, int b) {
        return (a / gcd(a, b)) * b;
    }

    static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    static class Edge implements Comparable<Edge> {
        int u, v, w;

        Edge(int a, int b, int c) {
            u = a;
            v = b;
            w = c;
        }

        public int compareTo(Edge e) {
            return w - e.w;
        }
    }

    static class Pair implements Comparable<Pair> {
        Long a, b;
        int c;

        public Pair(Long a, Long b) {
            this.a = a;
            this.b = b;
//            this.c = c;
        }

        public String toString() {
            return a + " " + b + " ";
        }

        @Override
        public int compareTo(Pair o) {
            return (int) (a - o.a);
        }
    }

    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String file) throws IOException {
            br = new BufferedReader(new FileReader(file));
        }

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String readAllLines(BufferedReader reader) throws IOException {
            StringBuilder content = new StringBuilder();
            String line;
            while ((line = reader.readLine()) != null) {
                content.append(line);
                content.append(System.lineSeparator());
            }
            return content.toString();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec) f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongarr(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongarr(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntarr(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerarr(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

    }
}

**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 expose the bug in the given code, we need to focus on edge cases or scenarios that might cause the code to fail. The code seems to handle the main logic of calculating the minimum operations required to form a subarray of length `k` with a sliding window approach. However, there are a few potential issues that could be exploited:

1. **Edge Cases with Small Arrays**: Test cases where the array length `n` is very small (e.g., 1 or 2) and `k` is equal to `n`.
2. **Edge Cases with Large Arrays**: Test cases where `n` and `q` are at their maximum limits.
3. **Edge Cases with Specific Values**: Test cases where the array values are such that the sliding window logic might fail.
4. **Edge Cases with Repeated Values**: Test cases where the array has repeated values that might cause issues with the frequency map and tree set.

Let's create a small-scale test input that is complete and valid to expose the bug:

### Test Input
```plaintext
1
5 3 2
1 2 3 4 5
1 3
2 4
```

### Explanation
- **Test Case 1**:
  - `n = 5`, `k = 3`, `q = 2`
  - Array: `1 2 3 4 5`
  - Queries:
    - `1 3`: Subarray `[1, 2, 3]`
    - `2 4`: Subarray `[2, 3, 4]`

### Expected Output
- For the first query `[1, 2, 3]`, the minimum number of operations required to form a subarray of length 3 is 0.
- For the second query `[2, 3, 4]`, the minimum number of operations required to form a subarray of length 3 is 0.

### Why This Test Input?
- **Small Array**: The array length is small (5), making it easier to manually verify the correctness of the output.
- **Consecutive Values**: The array values are consecutive, which should not require any operations to form a subarray of length `k`.
- **Multiple Queries**: Two queries to ensure the sliding window logic is correctly implemented.

This test input should help identify any issues with the sliding window and frequency map logic 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: Yunli's Subarray Queries (easy version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $$$r=l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$. Note that in this version, you are only required to output $$$f([a_l, a_{l+1}, \ldots, a_{l+k-1}])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r=l+k-1$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the first query of the first testcase, $$$b=[1,2,3,2,1]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$2$$$ moves:

- Set $$$b_4=4$$$
- Set $$$b_5=5$$$

In the second query of the first testcase, $$$b=[2,3,2,1,2]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$3$$$ moves:

- Set $$$b_3=0$$$
- Set $$$b_2=-1$$$
- Set $$$b_1=-2$$$

Buggy Code:
import java.util.*;
import java.util.stream.Collectors;

public class YunelProplems {

     static int getNumberOfStepToSort(List<Integer> arr){
        int numberOfTrials, minNumberOfTrials = 0;
        for (int i = 1; i < arr.size();i++){
            numberOfTrials = 0;
            int temp = arr.get(i);
            for (int j = i - 1; j >= 0;j--){
                if(arr.get(j) == temp - 1) {
                    temp = temp - 1;
                    continue;
                }
                numberOfTrials++;
                if (j > 0) temp = temp - 1;
            }

            if (i != arr.size() - 1) {
                temp = arr.get(i);
                for (int j = i + 1; j < arr.size(); j++) {
                    if (arr.get(j) == temp + 1) {
                        temp = temp + 1;
                        i = j - 1 ;
                        continue;
                    }
                    numberOfTrials++;
                    temp = temp + 1;
                }
            }
            if (minNumberOfTrials == 0) {
                minNumberOfTrials = numberOfTrials;
            } else if(minNumberOfTrials > numberOfTrials){
                minNumberOfTrials = numberOfTrials;
            }
        }
         return minNumberOfTrials;
     }

    static Scanner s = new Scanner(System.in);
    public static void main(String[] args) {
        int testcases = s.nextInt();
        List<Integer> result = new ArrayList<>();
        for (int o = 0; o < testcases; o++) {
            int n = s.nextInt();
            int k = s.nextInt();
            int q = s.nextInt();
            s.nextLine();
            String input = s.nextLine();
            List<Integer> arr = Arrays.stream(input.split(" "))
                    .map(Integer::valueOf)
                    .collect(Collectors.toList());
            int[][] trials = new int[q][2];
            for (int i = 0; i < q; i++) {
                trials[i][0] = s.nextInt();
                trials[i][1] = s.nextInt();
            }

            for (int i = 0; i < q; i++) {
                List<Integer> arr1 = arr.subList(trials[i][0] - 1, trials[i][1]);
                result.add(getNumberOfStepToSort(arr1));
            }
        }

    result.forEach(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:
To generate a challenging test input that exposes the bug in the provided 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 seems to have issues with correctly calculating the minimum number of operations required to form a consecutive subarray of length `k`. It iterates through the array and tries to form a sequence, but it may not handle cases where the array elements are not in a strictly increasing or decreasing sequence efficiently.

2. **Edge Cases**:
   - **Small Arrays**: Test with the smallest possible arrays to ensure the code handles small inputs correctly.
   - **Large Arrays**: Test with the largest possible arrays to ensure the code handles large inputs efficiently.
   - **Uniform Arrays**: Test with arrays where all elements are the same.
   - **Sequences with Gaps**: Test with arrays that have large gaps between elements.
   - **Reversed Sequences**: Test with arrays that are in reverse order.
   - **Mixed Sequences**: Test with arrays that have both increasing and decreasing sequences.

3. **Challenging Scenarios**:
   - ** Queries with Small `k`**: Test with small values of `k` to ensure the code can handle short subarrays.
   - ** Queries with Large `k`**: Test with large values of `k` to ensure the code can handle long subarrays.
   - ** Queries with `l` and `r` Close to Boundaries**: Test with queries where `l` and `r` are close to the boundaries of the array.

Based on these considerations, here is a small-scale test input that is complete and valid:

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

This test input includes:
- A small array with a small `k` and two queries.
- A larger array with a larger `k` and three queries.
- An array in reverse order with a small `k` and two queries.

These scenarios 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: Yunli's Subarray Queries (easy version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $$$r=l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$. Note that in this version, you are only required to output $$$f([a_l, a_{l+1}, \ldots, a_{l+k-1}])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r=l+k-1$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the first query of the first testcase, $$$b=[1,2,3,2,1]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$2$$$ moves:

- Set $$$b_4=4$$$
- Set $$$b_5=5$$$

In the second query of the first testcase, $$$b=[2,3,2,1,2]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$3$$$ moves:

- Set $$$b_3=0$$$
- Set $$$b_2=-1$$$
- Set $$$b_1=-2$$$

Buggy Code:
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")
import sys
from itertools import permutations
from heapq import *
from collections import deque
import random
import bisect
from math import gcd

mi = lambda :map(int,input().split())
li = lambda :list(mi())

INF = 10**18
from collections import Counter
for i in range(int(input())):
    n,k,q=mi()
    a=li()
    d={}
    ans=[0]*(n-k+1)
    for i in range(k):
        d[a[i]-i]=d.get(a[i]-i,0)+1
    import heapq as hp
    h=[]
    for i in d.values():
        hp.heappush(h,-i)
    ans[0]=k+h[0]

    #print(d,ans)

    for i in range(k,n):
        i1=i-k
        if h[0]==-d[a[i1]-i1]:
            hp.heappop(h)
            hp.heappush(h,-d[a[i1]-i1]+1)
        d[a[i1]-i1]-=1
        if a[i]-i in d and h[0]==-d[a[i]-i]:
            hp.heappop(h)
        d[a[i]-i]=d.get(a[i]-i,0)+1
        hp.heappush(h,-d[a[i]-i])
        #print(d,h)
        ans[i-k+1]=k+h[0]
    for i in range(q):
        l,r=mi()
        print(ans[l-1])

**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 3 2
1 2 3 4 5
1 3
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (easy version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $$$r=l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$. Note that in this version, you are only required to output $$$f([a_l, a_{l+1}, \ldots, a_{l+k-1}])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r=l+k-1$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the first query of the first testcase, $$$b=[1,2,3,2,1]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$2$$$ moves:

- Set $$$b_4=4$$$
- Set $$$b_5=5$$$

In the second query of the first testcase, $$$b=[2,3,2,1,2]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$3$$$ moves:

- Set $$$b_3=0$$$
- Set $$$b_2=-1$$$
- Set $$$b_1=-2$$$

Buggy Code:
import sys
from collections import defaultdict,deque,Counter
from heapq import heappush,heappop,heapify
from math import inf,comb,ceil,log
from functools import cache
from itertools import accumulate,combinations,permutations
from bisect import bisect_left,bisect_right,insort_left
input=lambda:sys.stdin.readline().rstrip()
inp=lambda:map(int,input().split())
t=int(input())
def adv(x):
    e=a[x]
    occ[e]+=1
    heappush(h,(-occ[e],e))
def ret(x):
    e=a[x]
    occ[e]-=1
    heappush(h,(-occ[e],e))
for _ in range(t):
    n,k,q=inp()
    *a,=inp()
    a=[a[i]-i for i in range(n)]
    h=[]
    occ=Counter()
    Q=[]
    for _ in range(q):
        l,r=inp()
        l-=1
        r-=1
        Q+=(l,r),
    L=ceil(len(Q)**.5)
    Q.sort(key=lambda t:(t[0]//L,t[1]))
    x,y=0,-1
    #print(a)
    for l,r in Q:
        #print(l,r)
        while y<r:
            adv(y+1)
            y+=1
            #print(occ)
        while y>r:
            ret(y)
            y-=1
        while x<l:
            ret(x)
            x+=1
        while x>l:
            adv(x-1)
            x-=1
        while h and occ[h[0][1]]!=-h[0][0]:heappop(h)
        #print(occ,a,x,y,l,r)
        print(k+h[0][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
2
5 3 2
1 2 3 4 5
1 3
2 4
10 5 3
1 2 3 4 5 6 7 8 9 10
1 5
2 6
3 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (easy version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $$$r=l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$. Note that in this version, you are only required to output $$$f([a_l, a_{l+1}, \ldots, a_{l+k-1}])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r=l+k-1$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the first query of the first testcase, $$$b=[1,2,3,2,1]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$2$$$ moves:

- Set $$$b_4=4$$$
- Set $$$b_5=5$$$

In the second query of the first testcase, $$$b=[2,3,2,1,2]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$3$$$ moves:

- Set $$$b_3=0$$$
- Set $$$b_2=-1$$$
- Set $$$b_1=-2$$$

Buggy Code:
#include<stdio.h>

int dp(int a[],int b[],int l,int r){
	b[l]=1;
	int max=1;
	for(int x=l+1;x<=r;x++){
		if(a[x]-a[x-1]==1){
			b[x]=b[x-1]+1;
		}
		else{
		    b[x]=1;
		    for(int y=l;y<x-1;y++){
			    if(a[x]-a[y]==(x-y)){
				    b[x]=(b[x]<b[y]+1)?b[y]+1:b[x];
			    }
		    }
		}

		if(max<b[x]){
			max=b[x];
		}
	}
	return max;
}

int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n,k,q;
		scanf("%d%d%d",&n,&k,&q);
		int a[n+1];
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		for(int i=1;i<=q;i++){
			int l,r;
		    int b[n+1];
			scanf("%d%d",&l,&r);
			int max=dp(a,b,l,r);
			printf("%d\n",k-max);
		}
	}
	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
1
5 3 1
3 2 1 4 5
1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (easy version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $$$r=l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$. Note that in this version, you are only required to output $$$f([a_l, a_{l+1}, \ldots, a_{l+k-1}])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r=l+k-1$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the first query of the first testcase, $$$b=[1,2,3,2,1]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$2$$$ moves:

- Set $$$b_4=4$$$
- Set $$$b_5=5$$$

In the second query of the first testcase, $$$b=[2,3,2,1,2]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$3$$$ moves:

- Set $$$b_3=0$$$
- Set $$$b_2=-1$$$
- Set $$$b_1=-2$$$

Buggy Code:
#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

int main() {
    int t;
    cin >> t;
    for (int test = 0; test < t; ++test) {
        int n, k, q;
        cin >> n >> k >> q;
        vector<int> a(n + 1);
        for (int i = 1; i <= n; ++i) {
            cin >> a[i];
        }
        for (int query = 0; query < q; ++query) {
            int l, r;
            cin >> l >> r;
            unordered_map<int, int> freq;
            for (int i = l; i <= r; ++i) {
                int diff = a[i] - (i - l);
                freq[diff]++;
            }
            int max_freq = 0;
            for (auto& pair : freq) {
                if (pair.second > max_freq) {
                    max_freq = pair.second;
                }
            }
            cout << (k - max_freq) << endl;
        }
    }
    return 0;
}

// MGqmdKau5xNpjAYTyXw

**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 3 1
1 2 3 4 5
1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (easy version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $$$r=l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$. Note that in this version, you are only required to output $$$f([a_l, a_{l+1}, \ldots, a_{l+k-1}])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r=l+k-1$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the first query of the first testcase, $$$b=[1,2,3,2,1]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$2$$$ moves:

- Set $$$b_4=4$$$
- Set $$$b_5=5$$$

In the second query of the first testcase, $$$b=[2,3,2,1,2]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$3$$$ moves:

- Set $$$b_3=0$$$
- Set $$$b_2=-1$$$
- Set $$$b_1=-2$$$

Buggy Code:
import java.util.*;
import java.util.stream.Collectors;

public class YunelProplems {

     static int getNumberOfStepToSort(List<Integer> arr){
        int numberOfTrials, minNumberOfTrials = 0;
        for (int i = 1; i < arr.size();i++){
            numberOfTrials = 0;
            int temp = arr.get(i);
            for (int j = i - 1; j >= 0;j--){
                if(arr.get(j) == temp - 1) {
                    temp = temp - 1;
                    continue;
                }
                numberOfTrials++;
                if (j > 0) temp = temp - 1;
            }
            if (i != arr.size() - 1) {
                temp = arr.get(i);
                for (int j = i + 1; j < arr.size(); j++) {
                    if (arr.get(j) == temp + 1) {
                        temp = temp + 1;
                        continue;
                    }
                    numberOfTrials++;
                    temp = temp + 1;
                }
            }
            if (minNumberOfTrials == 0) {
                minNumberOfTrials = numberOfTrials;
            } else if(minNumberOfTrials > numberOfTrials){
                minNumberOfTrials = numberOfTrials;
            }
        }
         return minNumberOfTrials;
     }

    static Scanner s = new Scanner(System.in);
    public static void main(String[] args) {
        int testcases = s.nextInt();
        List<Integer> result = new ArrayList<>();
        for (int o = 0; o < testcases; o++) {
            int n = s.nextInt();
            int k = s.nextInt();
            int q = s.nextInt();
            s.nextLine();
            String input = s.nextLine();
            List<Integer> arr = Arrays.stream(input.split(" "))
                    .map(Integer::valueOf)
                    .collect(Collectors.toList());
            int[][] trials = new int[q][2];
            for (int i = 0; i < q; i++) {
                trials[i][0] = s.nextInt();
                trials[i][1] = s.nextInt();
            }

            for (int i = 0; i < q; i++) {
                List<Integer> arr1 = arr.subList(trials[i][0] - 1, trials[i][1]);
                result.add(getNumberOfStepToSort(arr1));
            }
        }

    result.forEach(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
2
5 3 1
1 2 3 4 5
1 3
5 3 1
5 4 3 2 1
1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (easy version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, it is guaranteed that $$$r=l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$. Note that in this version, you are only required to output $$$f([a_l, a_{l+1}, \ldots, a_{l+k-1}])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r=l+k-1$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the first query of the first testcase, $$$b=[1,2,3,2,1]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$2$$$ moves:

- Set $$$b_4=4$$$
- Set $$$b_5=5$$$

In the second query of the first testcase, $$$b=[2,3,2,1,2]$$$. Yunli can make a consecutive subarray of length $$$5$$$ in $$$3$$$ moves:

- Set $$$b_3=0$$$
- Set $$$b_2=-1$$$
- Set $$$b_1=-2$$$

Buggy Code:
import java.util.*;
import java.util.stream.Collectors;

public class YunelProplems {

     static int getNumberOfStepToSort(List<Integer> arr){
        int numberOfTrials, minNumberOfTrials = 0;
        for (int i = 1; i < arr.size();i++){
            numberOfTrials = 0;
            int temp = arr.get(i);
            for (int j = i - 1; j >= 0;j--){
                if(arr.get(j) == temp - 1) {
                    temp = temp - 1;
                    continue;
                }
                numberOfTrials++;
                if (j > 0) temp = temp - 1;
            }

            if (i != arr.size() - 1) {
                temp = arr.get(i);
                for (int j = i + 1; j < arr.size(); j++) {
                    if (arr.get(j) == temp + 1) {
                        temp = temp + 1;
                        continue;
                    }
                    numberOfTrials++;
                    temp = temp + 1;
                }
            }
            if (minNumberOfTrials == 0) {
                minNumberOfTrials = numberOfTrials;
            } else if(minNumberOfTrials > numberOfTrials){
                minNumberOfTrials = numberOfTrials;
            }
        }
         return minNumberOfTrials;
     }

    static Scanner s = new Scanner(System.in);
    public static void main(String[] args) {
        int testcases = s.nextInt();
        List<Integer> result = new ArrayList<>();
        for (int o = 0; o < testcases; o++) {
            int n = s.nextInt();
            int k = s.nextInt();
            int q = s.nextInt();
            s.nextLine();
            String input = s.nextLine();
            List<Integer> arr = Arrays.stream(input.split(" "))
                    .map(Integer::valueOf)
                    .collect(Collectors.toList());
            int[][] trials = new int[q][2];
            for (int i = 0; i < q; i++) {
                trials[i][0] = s.nextInt();
                trials[i][1] = s.nextInt();
            }

            for (int i = 0; i < q; i++) {
                List<Integer> arr1 = arr.subList(trials[i][0] - 1, trials[i][1]);
                result.add(getNumberOfStepToSort(arr1));
            }
        }

    result.forEach(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:
To expose the bug in the given code, we need to focus on edge cases or scenarios where the algorithm might fail. The key aspects to consider are:

1. **Consecutive Subarrays of Length k**: Ensure the queries always have `r = l + k - 1`.
2. **Edge Cases**: Test with arrays that have repeated elements, arrays with all unique elements, and arrays with elements that are already in a consecutive subarray.
3. **Small Scale**: The test input should be small but effective in exposing the bug.

Given the code, the main issue seems to be in the `getNumberOfStepToSort` function, where it calculates the minimum number of operations needed to make a subarray consecutive. The function might not correctly handle cases where the subarray already has consecutive elements or where the elements are not in a strictly increasing or decreasing order.

Here's a small-scale test input that can expose the bug:

```plaintext
2
5 3 2
1 2 3 4 5
1 3
2 4
6 4 2
2 3 4 5 6 7
1 4
2 5
```

Explanation:
1. **Test Case 1**:
   - Array: `1 2 3 4 5`
   - Query 1: `1 3` (Subarray: `1 2 3`)
   - Query 2: `2 4` (Subarray: `2 3 4`)

2. **Test Case 2**:
   - Array: `2 3 4 5 6 7`
   - Query 1: `1 4` (Subarray: `2 3 4 5`)
   - Query 2: `2 5` (Subarray: `3 4 5 6`)

In these test cases, the subarrays are already consecutive, and the expected output for each query should be `0`. However, the given code might not handle this correctly and might return a non-zero value, exposing the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chamo and Mocha's Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Chamo gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha doesn't like arrays containing different numbers, so Mocha decides to use magic to change the array. Mocha can perform the following three-step operation some (possibly, zero) times:

  1. Choose indices $l$ and $r$ ($1 \leq l < r \leq n$)    2. Let $x$ be the median$^\dagger$ of the subarray $[a_l, a_{l+1},\ldots, a_r]$    3. Set all values $a_l, a_{l+1},\ldots, a_r$ to $x$

Suppose $a=[1,2,3,4,5]$ initially:

  * If Mocha chooses $(l,r)=(3,4)$ in the first operation, then $x=3$, the array will be changed into $a=[1,2,3,3,5]$.    * If Mocha chooses $(l,r)=(1,3)$ in the first operation, then $x=2$, the array will be changed into $a=[2,2,2,4,5]$.

Mocha will perform the operation until the array contains only the same number. Mocha wants to know what is the maximum possible value of this number.

$^\dagger$ The median in an array $b$ of length $m$ is an element that occupies position number $\lfloor \frac{m+1}{2} \rfloor$ after we sort the elements in non-decreasing order. For example, the median of $[3,1,4,1,5]$ is $3$ and the median of $[5,25,20,24]$ is $20$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output the maximum value of the number.

Examples:
input:
2
2
1 2
5
1 2 3 4 5
output:
1
4

Note: In the first test case, $a=[1,2]$. Mocha can only choose the interval $(l,r)=(1,2)$. The array will be changed to $a=[1,1]$. Therefore, the answer is $1$.

In the second test case, Mocha can perform the following operations:

  * Choose the interval $(l,r)=(4,5)$, then $a=[1,2,3,4,4]$.    * Choose the interval $(l,r)=(3,5)$, then $a=[1,2,4,4,4]$.    * Choose the interval $(l,r)=(1,5)$, then $a=[4,4,4,4,4]$.

The array contains only the same number, which is $4$. It can be proven that the maximum value of the final number cannot be greater than $4$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
ll mod =(ll) 998244353;
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;

// #define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update>

ll power(ll a, ll b) {
    ll ans = 1;
    while (b != 0) {
      if (b % 2 == 1) {
      // if (ans>(LLONG_MAX)/a){
      //     return LLONG_MAX;
      // }
         ans = ans * a;
      }
      a = a * a;
      b = b / 2;
      ans%=mod;
      a%=mod;
    }
    return ans;
}
ll modInverse(ll A, ll M)
{
    return power(A,M-2);
}

ll gcd (ll x,ll y){
    ll b = min(x,y);
    ll a = max(x,y);
    if (b==0){
        return a;
    }
    else{
        return gcd(b,a%b);
    }
}

ll ceil_divide (ll a, ll b){
    ll ans = a/b;
    if(a%b!=0){
        ans++;
    }
    return ans;
}

// ll gcd_arr(ll arr[]){
//     ll n=arr.size();
//     ll g=arr[0];
//     for (ll i=1;i<n;i++){
//         g= gcd(max(g,arr[i]),min(g,arr[i]));
//     }
//     return g;
// }

// const ll N=1000000;
// vector<ll> V(N+1);

// //SeiveOfEratosthenes
// void sieve(){
//     for (ll i=1;i<=N;i++) V[i]=-1;
//     V[1]=-1;
//     for (ll i=2;i<=N;i++){
//         if (V[i]!=-1) continue;
//         for (ll j=2*i;j<=N;j+=i){
//             V[j]=i;
//         }
//     }
// }

// bool isPrime(ll n){
//     if((n==1)||(V[n]!=-1)){
//         return false;
//     }
//     return true;
// }

// vector<ll> primeFactoristion (ll n){
//     vector<ll> pf;
//     while(V[n]!=-1){
//         pf.push_back(V[n]);
//         n=n/
//         V[n];
//     }
//     pf.push_back(n);
//     reverse(pf.begin(),pf.end());
//     return pf;
// }

vector<ll> divisors(ll n){
    vector<ll> div;
    if(n==1){
        div.push_back(1);
        return div;
    }
    for(ll i=1;i<=(sqrtl(n));i++){
        if((n%i)==0){
            div.push_back(i);
            div.push_back(n/i);
        }
    }
    sort(div.begin(),div.end());
    return div;
}

// vector<ll> factorial(N);

// void fac(){
//     factorial[0]=1;
//     for(int i=1;i<N+1;i++){
//         factorial[i]=(factorial[p.x-1]*p.y)%mod;
//     }
// }

// ll ncr(ll n, ll r){
//     return factorial[n]/(factorial[n-r]*factorial[r]);
// }

// vector<ll> adj[200001];
// ll visited[200001];
// ll dp[200001][2];
// ll n;

// int no_of_digits(ll x){
//     ll c =0;
//     while(x!=0){
//         c++;
//         x/=10;
//     }
//     return c;

// // }
// vector<ll> func(ll l, ll r){
//     ll length = r-l+1;
//     vector<pair<ll,ll>> v1,v2;
//     while(length != 0){
//         ll k = 1;
//         ll i_max = -1;
//         while(l%k==0){
//             i_max++;
//             k*=2;
//         }
//         k/=2;
//         // cout<<k<<'k';
//         ll l_max = -1;
//         ll kk = 1;
//         while(kk<=length){
//             l_max++;
//             kk*=2;
//         }
//         kk/=2;
//         ll l1 = min(i_max,l_max);
//         ll r1 = l/min(k,kk);
//         v.push_back({l1,r1});
//         length -= min(kk,k);
//         l = min(k,kk)*(r1+1);
//         length = r-l+1;
//         cout<<l1<<' '<<r1<<' '<<l<<'l'<<length<<'\n';
//     }
//     return v1
// }

void solve () {
    ll n;
    cin>>n;
    ll arr[n];
    ll m = INT_MIN;
    ll m2;
    for(ll i=0;i<n;i++){
        cin>>arr[i];
        if(arr[i]>m){
            m2 = m;
            m = arr[i];
        }
    }
    ll j = -3;
    for(ll i=0;i<n;i++){
        if(arr[i]==m){
            if(i-j>2){
                j = i;
            }
            else{
                cout<<m<<'\n';
                return;
            }
        }
    }
    cout<<m2<<'\n';

}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    ll t = 1;
    cin>>t;
    // sieve();
    while(t--){
        solve();
    }
    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
3
5 5 5
4
1 2 3 4
5
1 3 2 5 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chamo and Mocha's Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Chamo gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha doesn't like arrays containing different numbers, so Mocha decides to use magic to change the array. Mocha can perform the following three-step operation some (possibly, zero) times:

  1. Choose indices $l$ and $r$ ($1 \leq l < r \leq n$)    2. Let $x$ be the median$^\dagger$ of the subarray $[a_l, a_{l+1},\ldots, a_r]$    3. Set all values $a_l, a_{l+1},\ldots, a_r$ to $x$

Suppose $a=[1,2,3,4,5]$ initially:

  * If Mocha chooses $(l,r)=(3,4)$ in the first operation, then $x=3$, the array will be changed into $a=[1,2,3,3,5]$.    * If Mocha chooses $(l,r)=(1,3)$ in the first operation, then $x=2$, the array will be changed into $a=[2,2,2,4,5]$.

Mocha will perform the operation until the array contains only the same number. Mocha wants to know what is the maximum possible value of this number.

$^\dagger$ The median in an array $b$ of length $m$ is an element that occupies position number $\lfloor \frac{m+1}{2} \rfloor$ after we sort the elements in non-decreasing order. For example, the median of $[3,1,4,1,5]$ is $3$ and the median of $[5,25,20,24]$ is $20$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output the maximum value of the number.

Examples:
input:
2
2
1 2
5
1 2 3 4 5
output:
1
4

Note: In the first test case, $a=[1,2]$. Mocha can only choose the interval $(l,r)=(1,2)$. The array will be changed to $a=[1,1]$. Therefore, the answer is $1$.

In the second test case, Mocha can perform the following operations:

  * Choose the interval $(l,r)=(4,5)$, then $a=[1,2,3,4,4]$.    * Choose the interval $(l,r)=(3,5)$, then $a=[1,2,4,4,4]$.    * Choose the interval $(l,r)=(1,5)$, then $a=[4,4,4,4,4]$.

The array contains only the same number, which is $4$. It can be proven that the maximum value of the final number cannot be greater than $4$.

Buggy Code:
#include <bits/stdc++.h>

using namespace std;
#define ll long long

int main() {
    int t;
    cin >> t;

    while (t--) {
        ll n;
        cin >> n;

        vector<ll> arr(n);
        for (ll i = 0; i < n; i++) {
            cin >> arr[i];
        }
        sort(arr.begin(),arr.end());
        cout<<arr[n-2]<<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:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the problem and the faulty implementation:

### Problem Analysis:
- The goal is to find the maximum possible value in the array after performing the operation multiple times.
- The operation involves choosing a subarray and setting all elements in the subarray to the median of the subarray.
- The median of a subarray is the middle element after sorting the subarray.

### Buggy Code Analysis:
- The buggy code sorts the entire array and outputs the second largest element (`arr[n-2]`).
- This approach is incorrect because the median of a subarray is not necessarily the second largest element in the sorted array.

### Edge Cases to Consider:
1. **Array with all elements the same**: This is a trivial case where the output should be the same element.
2. **Array with a single unique maximum value**: The maximum value should be the output.
3. **Array with a mix of values where the median of a subarray is not the second largest element**: This is a critical case to expose the bug.

### Challenging Test Input:
- Let's consider an array where the median of a subarray is not the second largest element in the sorted array.
- An example is an array with a mix of values where the median of a subarray is the largest element.

### Test Input:
```plaintext
1
5
1 3 5 4 2
```

### Explanation:
- The array is `[1, 3, 5, 4, 2]`.
- If Mocha performs the operation on the entire array, the median is `3` (the middle element after sorting `[1, 2, 3, 4, 5]`).
- However, if Mocha performs the operation on the subarray `[3, 5, 4]`, the median is `4` (the middle element after sorting `[3, 4, 5]`).
- The correct output should be `4`, but the buggy code will output `5` (the second largest element in the sorted array).

This test input should expose the bug 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: Chamo and Mocha's Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Chamo gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha doesn't like arrays containing different numbers, so Mocha decides to use magic to change the array. Mocha can perform the following three-step operation some (possibly, zero) times:

  1. Choose indices $l$ and $r$ ($1 \leq l < r \leq n$)    2. Let $x$ be the median$^\dagger$ of the subarray $[a_l, a_{l+1},\ldots, a_r]$    3. Set all values $a_l, a_{l+1},\ldots, a_r$ to $x$

Suppose $a=[1,2,3,4,5]$ initially:

  * If Mocha chooses $(l,r)=(3,4)$ in the first operation, then $x=3$, the array will be changed into $a=[1,2,3,3,5]$.    * If Mocha chooses $(l,r)=(1,3)$ in the first operation, then $x=2$, the array will be changed into $a=[2,2,2,4,5]$.

Mocha will perform the operation until the array contains only the same number. Mocha wants to know what is the maximum possible value of this number.

$^\dagger$ The median in an array $b$ of length $m$ is an element that occupies position number $\lfloor \frac{m+1}{2} \rfloor$ after we sort the elements in non-decreasing order. For example, the median of $[3,1,4,1,5]$ is $3$ and the median of $[5,25,20,24]$ is $20$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output the maximum value of the number.

Examples:
input:
2
2
1 2
5
1 2 3 4 5
output:
1
4

Note: In the first test case, $a=[1,2]$. Mocha can only choose the interval $(l,r)=(1,2)$. The array will be changed to $a=[1,1]$. Therefore, the answer is $1$.

In the second test case, Mocha can perform the following operations:

  * Choose the interval $(l,r)=(4,5)$, then $a=[1,2,3,4,4]$.    * Choose the interval $(l,r)=(3,5)$, then $a=[1,2,4,4,4]$.    * Choose the interval $(l,r)=(1,5)$, then $a=[4,4,4,4,4]$.

The array contains only the same number, which is $4$. It can be proven that the maximum value of the final number cannot be greater than $4$.

Buggy Code:
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define pob pop_back
#define F first
#define S second
#define nl '\n'
#define lb lower_bound
#define ub upper_bound
#define read(a)			\
	for(auto &i : a)	\
		cin>>i;
#define ot(a)			\
	for(auto &i : a)    \
		cout<<i<<" ";   \
	cout << "\n";
typedef long long ll;
#define int ll
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;
typedef vector<pair<int,int>> vpll;
const int mod = 1000000007;
const int N = 200001;
const int no = 998244353;
#define inf (int)((int)1e18+2)

void solve(){
	int n;
	cin >> n;
	vi a(n+1);
	a[0]=-inf;
	for(int i=1; i<=n; i++){
		cin >> a[i];
	}
	int ans=0;
	for(int i=1; i<n; i++){
		ans=max(ans,min(a[i],a[i+1]));
	}
	for(int i=0; i<n-2; i++){
		if(a[i]==a[i+1] || a[i]==a[i+2]) ans=max(ans,a[i]);
		else if(a[i+1]==a[i+2]) ans=max(ans,a[i+1]);
		else{
			vi v={a[i],a[i+1],a[i+2]};
			sort(all(v));
			ans=max(ans,v[1]);
		}
	}
	cout << ans << nl;

	// cerr<<"worked in "<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms\n";
}

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int testCases;
    cin >> testCases;
    while(testCases--)
    	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
2
3
3 3 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: Chamo and Mocha's Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Chamo gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha doesn't like arrays containing different numbers, so Mocha decides to use magic to change the array. Mocha can perform the following three-step operation some (possibly, zero) times:

  1. Choose indices $l$ and $r$ ($1 \leq l < r \leq n$)    2. Let $x$ be the median$^\dagger$ of the subarray $[a_l, a_{l+1},\ldots, a_r]$    3. Set all values $a_l, a_{l+1},\ldots, a_r$ to $x$

Suppose $a=[1,2,3,4,5]$ initially:

  * If Mocha chooses $(l,r)=(3,4)$ in the first operation, then $x=3$, the array will be changed into $a=[1,2,3,3,5]$.    * If Mocha chooses $(l,r)=(1,3)$ in the first operation, then $x=2$, the array will be changed into $a=[2,2,2,4,5]$.

Mocha will perform the operation until the array contains only the same number. Mocha wants to know what is the maximum possible value of this number.

$^\dagger$ The median in an array $b$ of length $m$ is an element that occupies position number $\lfloor \frac{m+1}{2} \rfloor$ after we sort the elements in non-decreasing order. For example, the median of $[3,1,4,1,5]$ is $3$ and the median of $[5,25,20,24]$ is $20$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output the maximum value of the number.

Examples:
input:
2
2
1 2
5
1 2 3 4 5
output:
1
4

Note: In the first test case, $a=[1,2]$. Mocha can only choose the interval $(l,r)=(1,2)$. The array will be changed to $a=[1,1]$. Therefore, the answer is $1$.

In the second test case, Mocha can perform the following operations:

  * Choose the interval $(l,r)=(4,5)$, then $a=[1,2,3,4,4]$.    * Choose the interval $(l,r)=(3,5)$, then $a=[1,2,4,4,4]$.    * Choose the interval $(l,r)=(1,5)$, then $a=[4,4,4,4,4]$.

The array contains only the same number, which is $4$. It can be proven that the maximum value of the final number cannot be greater than $4$.

Buggy Code:
# ﷽
import sys
input = lambda: sys.stdin.readline().strip()
def inlst():return [int(i) for i in input().split()]
oo=float('inf')

def solve():
    n=int(input())
    lst=inlst()
    pref=[0]*(n+1)
    suff=[0]*(n+1)
    pref[1]=0;suff[n-1]=n-1
    for i in range(1,n):
        if lst[i]>lst[i-1]:
            pref[i+1]=i
        else:
            pref[i+1]=pref[i]
    for i in range(n-2,-1,-1):
        if lst[i]>lst[i+1]:
            suff[i]=i
        else:
            suff[i]=suff[i+1]
    enm=sorted(enumerate(lst),key=lambda x:x[1])
    ans=sorted(lst)[(n-1)//2]
    for i,j in  enm:
        idx=pref[i]
        if lst[idx]>=j:
            abs(idx-i)<=3
            ans=max(ans,j)
        idx=suff[i+1]
        if lst[idx]>=j:
            abs(i-idx)<=3
            ans=max(ans,j)

    print(ans)

def main():
    for i in range(int(input())):
        solve()

if __name__ == "__main__":
    main()

**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 expose the bug in the provided code, we need to focus on scenarios where the median selection and the subsequent operations might lead to unexpected results. The code seems to have issues with correctly handling the median and the ranges for which it performs the operations.

### Step-by-Step Analysis:

1. **Understanding the Bug:**
   - The code uses `pref` and `suff` arrays to keep track of the nearest indices where the values are greater or less than the current value.
   - The logic for updating `ans` is flawed because it does not correctly handle the median selection and the range checks.

2. **Edge Cases:**
   - **Single Element Array:** This is not possible as \( n \geq 2 \).
   - **Array with All Elements Equal:** This should not cause any issues.
   - **Array with Alternating Values:** This can expose issues in the median selection.
   - **Array with a Single Median Value:** This can test the logic for selecting the median.
   - **Array with a Small Range of Values:** This can test the range checks.

3. **Test Input:**
   - We need an array where the median is not straightforward to compute, and the operations might lead to incorrect results.

### Chosen Test Input:
```plaintext
2
5
1 3 2 4 5
4
1 2 4 3
```

### Explanation:
1. **First Test Case:**
   - Array: `[1, 3, 2, 4, 5]`
   - Median: `3` (after sorting: `[1, 2, 3, 4, 5]`)
   - The code should correctly identify `3` as the median and update the array accordingly. However, if the code incorrectly handles the ranges or the median selection, it might fail to identify `3` as the maximum possible value.

2. **Second Test Case:**
   - Array: `[1, 2, 4, 3]`
   - Median: `2` (after sorting: `[1, 2, 3, 4]`)
   - The code should correctly identify `2` as the median and update the array accordingly. However, if the code incorrectly handles the ranges or the median selection, it might fail to identify `2` as the maximum possible value.

### Expected Output:
```plaintext
3
2
```

This test input should help expose any issues in the median selection and range checks in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chamo and Mocha's Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Chamo gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha doesn't like arrays containing different numbers, so Mocha decides to use magic to change the array. Mocha can perform the following three-step operation some (possibly, zero) times:

  1. Choose indices $l$ and $r$ ($1 \leq l < r \leq n$)    2. Let $x$ be the median$^\dagger$ of the subarray $[a_l, a_{l+1},\ldots, a_r]$    3. Set all values $a_l, a_{l+1},\ldots, a_r$ to $x$

Suppose $a=[1,2,3,4,5]$ initially:

  * If Mocha chooses $(l,r)=(3,4)$ in the first operation, then $x=3$, the array will be changed into $a=[1,2,3,3,5]$.    * If Mocha chooses $(l,r)=(1,3)$ in the first operation, then $x=2$, the array will be changed into $a=[2,2,2,4,5]$.

Mocha will perform the operation until the array contains only the same number. Mocha wants to know what is the maximum possible value of this number.

$^\dagger$ The median in an array $b$ of length $m$ is an element that occupies position number $\lfloor \frac{m+1}{2} \rfloor$ after we sort the elements in non-decreasing order. For example, the median of $[3,1,4,1,5]$ is $3$ and the median of $[5,25,20,24]$ is $20$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output the maximum value of the number.

Examples:
input:
2
2
1 2
5
1 2 3 4 5
output:
1
4

Note: In the first test case, $a=[1,2]$. Mocha can only choose the interval $(l,r)=(1,2)$. The array will be changed to $a=[1,1]$. Therefore, the answer is $1$.

In the second test case, Mocha can perform the following operations:

  * Choose the interval $(l,r)=(4,5)$, then $a=[1,2,3,4,4]$.    * Choose the interval $(l,r)=(3,5)$, then $a=[1,2,4,4,4]$.    * Choose the interval $(l,r)=(1,5)$, then $a=[4,4,4,4,4]$.

The array contains only the same number, which is $4$. It can be proven that the maximum value of the final number cannot be greater than $4$.

Buggy Code:
# ^-^

def solve():
    n = int(input())
    l = list(map(int, input().split()))

    if n == 2:
        print(min(l[0], l[1]))
        return

    ans = l[1]
    for i in range(2, n-1):
        ans = max(ans, sorted([l[i-1], l[i], l[i+1]])[1])
    print(ans)

if __name__ == '__main__':
    t = 1
    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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider the following aspects:

1. **Edge Cases:**
   - Small arrays (e.g., `n = 2` or `n = 3`).
   - Arrays with all elements being the same.
   - Arrays with a few unique elements.

2. **Critical Scenarios:**
   - Arrays where the median of a subarray is not the middle element of the sorted subarray.
   - Arrays where the median changes significantly after each operation.

3. **Faulty Logic:**
   - The code only considers the median of subarrays of length 3, which is a significant limitation.
   - The code does not handle subarrays of length 2 correctly.

Given these considerations, let's construct a test case that will likely expose the bug:

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

### Explanation:
1. **Test Case 1:**
   - Array: `[1, 3, 2]`
   - The median of the entire array is `2`.
   - The code will only consider subarrays of length 3, which is the entire array in this case.
   - The correct answer is `2`.

2. **Test Case 2:**
   - Array: `[1, 4, 2, 3]`
   - The median of the entire array is `2.5` (which should be rounded to `2`).
   - The code will only consider subarrays of length 3, which are `[1, 4, 2]` and `[4, 2, 3]`.
   - The medians of these subarrays are `2` and `3` respectively.
   - The correct answer is `3`.

3. **Test Case 3:**
   - Array: `[1, 3, 5, 4, 2]`
   - The median of the entire array is `3`.
   - The code will only consider subarrays of length 3, which are `[1, 3, 5]`, `[3, 5, 4]`, and `[5, 4, 2]`.
   - The medians of these subarrays are `3`, `4`, and `4` respectively.
   - The correct answer is `4`.

These test cases should help expose the bug in the buggy code by highlighting the limitations of only considering subarrays of length 3 and the incorrect handling of subarrays of length 2.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chamo and Mocha's Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Chamo gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha doesn't like arrays containing different numbers, so Mocha decides to use magic to change the array. Mocha can perform the following three-step operation some (possibly, zero) times:

  1. Choose indices $l$ and $r$ ($1 \leq l < r \leq n$)    2. Let $x$ be the median$^\dagger$ of the subarray $[a_l, a_{l+1},\ldots, a_r]$    3. Set all values $a_l, a_{l+1},\ldots, a_r$ to $x$

Suppose $a=[1,2,3,4,5]$ initially:

  * If Mocha chooses $(l,r)=(3,4)$ in the first operation, then $x=3$, the array will be changed into $a=[1,2,3,3,5]$.    * If Mocha chooses $(l,r)=(1,3)$ in the first operation, then $x=2$, the array will be changed into $a=[2,2,2,4,5]$.

Mocha will perform the operation until the array contains only the same number. Mocha wants to know what is the maximum possible value of this number.

$^\dagger$ The median in an array $b$ of length $m$ is an element that occupies position number $\lfloor \frac{m+1}{2} \rfloor$ after we sort the elements in non-decreasing order. For example, the median of $[3,1,4,1,5]$ is $3$ and the median of $[5,25,20,24]$ is $20$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output the maximum value of the number.

Examples:
input:
2
2
1 2
5
1 2 3 4 5
output:
1
4

Note: In the first test case, $a=[1,2]$. Mocha can only choose the interval $(l,r)=(1,2)$. The array will be changed to $a=[1,1]$. Therefore, the answer is $1$.

In the second test case, Mocha can perform the following operations:

  * Choose the interval $(l,r)=(4,5)$, then $a=[1,2,3,4,4]$.    * Choose the interval $(l,r)=(3,5)$, then $a=[1,2,4,4,4]$.    * Choose the interval $(l,r)=(1,5)$, then $a=[4,4,4,4,4]$.

The array contains only the same number, which is $4$. It can be proven that the maximum value of the final number cannot be greater than $4$.

Buggy Code:
def s(c, i, a, j):
    t = 0
    k = j
    while 0 <= k < len(a):
        if c == 1 and k == len(a) - 1:
            return False
        if c == -1 and k == 0:
            return False
        k += c
        if a[k] >= i:
            t += 1
        else: t -= 1
        if t > 0:
            return True
    return False

def f(d, a, i):
    for j in d[i]:
        t = s(1, i, a, j) or s(-1, i, a, j)
        if t:
            return True

for _ in range(int(input())):
    n = int(input())
    a = [int(i) for i in input().split()]
    m = 0
    for i in range(len(a)):
        if i != 0 and a[i - 1] >= a[i]:
            m = max(m, a[i])
        if i != len(a) - 1 and a[i + 1] >= a[i]:
            m = max(m, a[i])
    print(m)

**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
2
3
1 3 2
4
3 1 2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chamo and Mocha's Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Chamo gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha doesn't like arrays containing different numbers, so Mocha decides to use magic to change the array. Mocha can perform the following three-step operation some (possibly, zero) times:

  1. Choose indices $l$ and $r$ ($1 \leq l < r \leq n$)    2. Let $x$ be the median$^\dagger$ of the subarray $[a_l, a_{l+1},\ldots, a_r]$    3. Set all values $a_l, a_{l+1},\ldots, a_r$ to $x$

Suppose $a=[1,2,3,4,5]$ initially:

  * If Mocha chooses $(l,r)=(3,4)$ in the first operation, then $x=3$, the array will be changed into $a=[1,2,3,3,5]$.    * If Mocha chooses $(l,r)=(1,3)$ in the first operation, then $x=2$, the array will be changed into $a=[2,2,2,4,5]$.

Mocha will perform the operation until the array contains only the same number. Mocha wants to know what is the maximum possible value of this number.

$^\dagger$ The median in an array $b$ of length $m$ is an element that occupies position number $\lfloor \frac{m+1}{2} \rfloor$ after we sort the elements in non-decreasing order. For example, the median of $[3,1,4,1,5]$ is $3$ and the median of $[5,25,20,24]$ is $20$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output the maximum value of the number.

Examples:
input:
2
2
1 2
5
1 2 3 4 5
output:
1
4

Note: In the first test case, $a=[1,2]$. Mocha can only choose the interval $(l,r)=(1,2)$. The array will be changed to $a=[1,1]$. Therefore, the answer is $1$.

In the second test case, Mocha can perform the following operations:

  * Choose the interval $(l,r)=(4,5)$, then $a=[1,2,3,4,4]$.    * Choose the interval $(l,r)=(3,5)$, then $a=[1,2,4,4,4]$.    * Choose the interval $(l,r)=(1,5)$, then $a=[4,4,4,4,4]$.

The array contains only the same number, which is $4$. It can be proven that the maximum value of the final number cannot be greater than $4$.

Buggy Code:
import javax.swing.plaf.IconUIResource;
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
import java.util.function.Supplier;

public class Main {

    static Input in = new Input();

    static long mod = (long) 1e9 + 7;

    static long inv(long p, long m) {

        long res = 1;
        while (m != 0) {
            if ((m & 1) == 1) res = res * p % mod;
            p = (p % mod) * (p % mod) % mod;
            m >>= 1;
        }

        return res;
    }

    public static void main(String[] args) throws IOException {
        //Scanner in = new Scanner(new File("input.txt"));//远古题专属
        //PrintWriter up = new PrintWriter(new File("output.txt"));//远古题专属
        Scanner in = new Scanner(System.in);
        PrintWriter up = new PrintWriter(System.out);
        Random ins = new Random();
        int t = in.nextInt();
        up:
        while (t-- > 0) {
            int n=in.nextInt();
            int[] s=new int[n];

            for(int i=0;i<n;i++){
                s[i]=in.nextInt();

            }
            int u=0;
            for(int i=1;i<n;i++) u=Math.max(u,Math.min(s[i],s[i-1]));
            up.println(u);

        }
        up.flush();
    }

    public static void dfs(List<List<int[]>> f,int x,int fu,int[][] hight,int[][] k2,int hp,int[] ph){

        ph[x]=hp;
        for(int[] d:f.get(x)){
            if(d[0]!=fu){
                k2[d[0]][0]=x;
                hight[d[0]][0]=d[1];
                for(int i=1;i<=20;i++){
                    k2[d[0]][i]=k2[k2[d[0]][i-1]][i-1];
                    hight[d[0]][i]=Math.min(hight[d[0]][i-1],hight[k2[d[0]][i-1]][i-1]);
                }
                dfs(f,d[0],x,hight,k2,hp+1,ph);
            }
        }
    }
    public static int lca(int[][] k2,int x,int y,int[] hp,int[][] hight){
        if(hp[x]<hp[y]){
            int u=x;
            x=y;y=u;
        }
        int u=10000000;
        for(int i=20;i>=0;i--){
            if(hp[k2[x][i]]>=hp[y]){
                u=Math.min(hight[x][i],u);
                x=k2[x][i];

            }
        }
        if(x==y) return u;

        for(int i=20;i>=0;i--){
            if(k2[x][i]!=k2[y][i]){
                u=Math.min(hight[x][i],u);
                u=Math.min(hight[y][i],u);
                x=k2[x][i];
                y=k2[y][i];

            }
        }

        u=Math.min(Math.min(hight[x][0],hight[y][0]),u);
        return u;
    }

    public static int gcd(int x, int y) {
        return y == 0 ? x : gcd(y, x % y);
    }

}

class node {
    node[] ne;
    int sum;

    public node() {
        ne=new node[27];
        sum=0;
    }
}

class Input {
    static StreamTokenizer streamTokenizer = new StreamTokenizer(System.in);

    //static BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
    static double nextDouble() throws IOException {
        streamTokenizer.nextToken();
        return (double) streamTokenizer.nval;
    }

    static int nextInt() throws IOException {
        streamTokenizer.nextToken();
        return (int) streamTokenizer.nval;
    }

    static long nextLong() throws IOException {
        streamTokenizer.nextToken();
        return (long) streamTokenizer.nval;
    }
}

**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 buggy code, we need to focus on edge cases and scenarios where the code's logic might fail. The buggy code seems to be using a simplistic approach to find the maximum possible value of the final number by iterating through the array and finding the maximum of the minimum values of adjacent pairs. This approach is incorrect because it does not consider the median of subarrays, which is a more complex operation.

### Key Points to Consider:
1. **Edge Cases:**
   - Arrays with all elements the same.
   - Arrays with a few unique elements.
   - Arrays where the median is not one of the boundary elements.

2. **Challenging Scenarios:**
   - Arrays where the median of subarrays is significantly different from the boundary elements.
   - Arrays with a large range of values but a small number of unique values.

### Test Input:
Let's create a test case that includes a mix of these scenarios:

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

### Explanation:
1. **Test Case 1:**
   - Array: `[1, 2, 3]`
   - The median of the subarray `[1, 2, 3]` is `2`. The final array can be `[2, 2, 2]`.
   - Correct output: `2`

2. **Test Case 2:**
   - Array: `[5, 1, 5, 1]`
   - The median of the subarray `[5, 1, 5, 1]` is `1`. The final array can be `[1, 1, 1, 1]`.
   - Correct output: `1`

3. **Test Case 3:**
   - Array: `[1, 3, 2, 4, 5]`
   - The median of the subarray `[1, 3, 2, 4, 5]` is `3`. The final array can be `[3, 3, 3, 3, 3]`.
   - Correct output: `3`

### Expected Output:
```plaintext
2
1
3
```

This test input should expose the bug in the provided code, as the code's logic of finding the maximum of the minimum values of adjacent pairs will not correctly handle the median of subarrays.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chamo and Mocha's Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Chamo gave her an array $a$ consisting of $n$ positive integers as a gift.

Mocha doesn't like arrays containing different numbers, so Mocha decides to use magic to change the array. Mocha can perform the following three-step operation some (possibly, zero) times:

  1. Choose indices $l$ and $r$ ($1 \leq l < r \leq n$)    2. Let $x$ be the median$^\dagger$ of the subarray $[a_l, a_{l+1},\ldots, a_r]$    3. Set all values $a_l, a_{l+1},\ldots, a_r$ to $x$

Suppose $a=[1,2,3,4,5]$ initially:

  * If Mocha chooses $(l,r)=(3,4)$ in the first operation, then $x=3$, the array will be changed into $a=[1,2,3,3,5]$.    * If Mocha chooses $(l,r)=(1,3)$ in the first operation, then $x=2$, the array will be changed into $a=[2,2,2,4,5]$.

Mocha will perform the operation until the array contains only the same number. Mocha wants to know what is the maximum possible value of this number.

$^\dagger$ The median in an array $b$ of length $m$ is an element that occupies position number $\lfloor \frac{m+1}{2} \rfloor$ after we sort the elements in non-decreasing order. For example, the median of $[3,1,4,1,5]$ is $3$ and the median of $[5,25,20,24]$ is $20$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 500$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2\leq n\leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^9$) — the elements of the array $a$.

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

Output: For each test case, output the maximum value of the number.

Examples:
input:
2
2
1 2
5
1 2 3 4 5
output:
1
4

Note: In the first test case, $a=[1,2]$. Mocha can only choose the interval $(l,r)=(1,2)$. The array will be changed to $a=[1,1]$. Therefore, the answer is $1$.

In the second test case, Mocha can perform the following operations:

  * Choose the interval $(l,r)=(4,5)$, then $a=[1,2,3,4,4]$.    * Choose the interval $(l,r)=(3,5)$, then $a=[1,2,4,4,4]$.    * Choose the interval $(l,r)=(1,5)$, then $a=[4,4,4,4,4]$.

The array contains only the same number, which is $4$. It can be proven that the maximum value of the final number cannot be greater than $4$.

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

public final class ProblemC {

    public static void solve(FastScanner in, PrintWriter out) throws Exception {
        int t = in.nextInt();
        for (int test = 0; test < t; test++) {
            int n = in.nextInt();
            int max = Integer.MIN_VALUE;
            for(int i = 0; i < n - 1; i++) {
                max = Math.max(max, in.nextInt());
            }
            int toss = in.nextInt();
            out.println(max);
        }
    }

    public static void main(String[] args) {
        try {
            final var in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
            final var out = new PrintWriter(System.out); // new BufferedOutputStream with byte printing might be faster
            solve(in, out);
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

}

class FastScanner {

    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }

    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }

}

**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
2
1 2
5
1 2 3 4 5
4
10 20 30 40
```