**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

Buggy Code:
//goal: hoa' cerberus (cho' 3 da`u)
//current state: cho' 0.5 da`u
//sub1: hydrorzshiba challenge (reward: cho' 0.5 da`u) COMPLETED
//sub2: expert codeforces (reward: ho'a cho'! (1 da`u))
//sub3: ptnk tst (reward: cho' 2 da`u)
//ULTIMATE SUB: VOI (reward: cerberus.)
//bonus: S or above PUPA MASTER 14.5 (reward: cho' + 0.25 da`u)

#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i = (a); i<=(int)(b); ++i)
#define FOD(i,a,b) for(int i = (a); i>=(int)(b); --i)
#define psb push_back
#define F first
#define int long long
#define S second
#define iall(a) (a).begin(), (a).end()
typedef long long ll;
#define nl '\n'

const int maxN = 2e5+5;
const int mod = 1e9+7;

/*
being retarded is exhausting
isn't it?
*/
/*
ll st[4*maxN];
int val[maxN], a[maxN];
int s[maxN], t[maxN], pos = 1;
int n,q;

vector<int> adj[maxN];

void dfs(int u, int par = -1) {
	s[u] = pos++;
	for (int v: adj[u]) {
		if (par == v) continue;
		dfs(v,u);
	}
	t[u] = pos-1;
}

void update(int u, int v, int id=1, int l=1, int r=n) {
	if (u < l || r < u) {
		return ;
    }
	if (l==r) {
		st[id] = 1LL*v;
		return;
	}
	int mid = (l+r) >> 1;
	if (u<=mid) update(u,v,2*id,l,mid);
	else update(u,v,2*id+1,mid+1,r);
	st[id] = st[2*id] + st[2*id+1];
}

ll query(int u, int v, int id=1, int l=1, int r=n) {
	if (l>v || r<u) return 0;
	if (l>=u && r<=v) return st[id];
	int mid = (l+r) >> 1;
	ll q1 = query(u, min(v,mid), 2*id, l, mid);
	ll q2 = query(max(u,mid+1), v, 2*id+1, mid+1, r);
	return q1+q2;
}
*/

void solve() {
	int n,k; cin >> n >> k;
	int a[n+1] = {}, b[n+1] {};
	FOR(i,1,n) cin >> a[i];
	FOR(i,1,n) cin >> b[i];
	priority_queue<pair<int,int>> q;
	ll ans = 0;
	FOR(i,1,n) q.push({a[i],i});
	while (k-- && !q.empty()) {
		int val = q.top().F, pos = q.top().S;
		q.pop();
		ans+=val;
		int newval = val-b[pos];
		if (newval>0) q.push({newval,pos});
	}
	cout << ans << nl;
}

int32_t main() {
	cin.tie(0) -> sync_with_stdio(0);
	int t=1; cin >> t;
	while (t--) solve();
}

//chu~`ng na`o zeena ms ho'a cho' 3 da`u?

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

Buggy Code:
import heapq
import sys

input = sys.stdin.read

def max_score(n, k, a, b):
    # Create a max-heap using negative values
    max_heap = []
    for i in range(n):
        heapq.heappush(max_heap, (-a[i], i))

    score = 0

    for _ in range(k):
        if not max_heap:
            break

        # Pop the maximum value (convert back to positive)
        max_val, index = heapq.heappop(max_heap)
        max_val = -max_val

        if max_val <= 0:
            break

        # Add it to the score
        score += max_val

        # Update the value in the array
        a[index] = max(0, a[index] - b[index])

        # If the updated value is greater than 0, push it back into the heap
        if a[index] > 0:
            heapq.heappush(max_heap, (-a[index], index))

    return score

def main():
    data = input().split()
    idx = 0
    t = int(data[idx])
    idx += 1
    results = []

    for _ in range(t):
        n = int(data[idx])
        k = int(data[idx + 1])
        idx += 2
        a = list(map(int, data[idx:idx + n]))
        idx += n
        b = list(map(int, data[idx:idx + n]))
        idx += n

        result = max_score(n, k, a, b)
        results.append(result)

    for result in results:
        print(result)

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 generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that might cause the code to fail. Here are the key points to consider:

1. **Initial Values and Reductions:**
   - The code uses a max-heap to always pick the largest value from array `a` and then reduces it by the corresponding value in array `b`.
   - If the reduction results in a value that is still positive, it is pushed back into the heap.

2. **Edge Cases:**
   - **Large `k` and Small `n`:** When `k` is very large and `n` is small, the code might run into issues if it doesn't handle the repeated reduction of the same elements efficiently.
   - **Zero or Negative Values:** The code should handle cases where the reduction results in zero or negative values correctly.
   - **Large Initial Values:** Large initial values in `a` and `b` can cause issues with integer overflow or large reductions.

3. **Test Case Generation:**
   - We will create a test case with a small `n` and a large `k` to ensure the code handles repeated operations correctly.
   - We will use values in `a` that are significantly larger than the corresponding values in `b` to ensure the heap is not empty prematurely.
   - We will also include a case where the reduction results in zero to ensure the code handles this correctly.

Here is a small-scale test input that should expose potential bugs in the code:

```plaintext
1
3 10
1000 1000 1000
900 900 900
```

Explanation:
- **Test Case 1:**
  - `n = 3` and `k = 10`
  - `a = [1000, 1000, 1000]`
  - `b = [900, 900, 900]`
  - The initial values in `a` are 1000, and the reduction values in `b` are 900. Each operation will reduce the value in `a` by 900, and the code should handle this reduction correctly over 10 operations.
  - The expected maximum score after 10 operations is `1000 + 100 + 100 + 100 + 10 + 10 + 10 + 1 + 1 + 1 = 1333`.

This test case should help identify any issues with the code's handling of repeated operations and reductions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

Buggy Code:
import heapq

def max_score(n, k, a, b):
    # Create a max-heap using negative values because Python has a min-heap by default
    heap = [(-a[i], b[i]) for i in range(n)]
    heapq.heapify(heap)

    score = 0

    while k > 0 and heap:
        # Pop the element with the highest value
        val, dec = heapq.heappop(heap)
        val = -val  # Convert back to positive

        # Add the value to the score
        score += val

        # Update the value and push it back if it's still positive
        new_val = max(0, val - dec)
        if new_val > 0:
            heapq.heappush(heap, (-new_val, dec))

        k -= 1

    return score

# Read the number of test cases
t = int(input())

results = []
for _ in range(t):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    results.append(max_score(n, k, a, b))

# Print the results for each test case
for result in results:
    print(result)

**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
5
3 3
1 2 3
3 2 1
5 5
10 9 8 7 6
1 2 3 4 5
1 1
1000000000
1000000000
10 10
1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
1 1000000000
1000000000
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

Buggy Code:
import java.util.*;

public class Scale {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        while (t-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            int[] a = new int[n];
            int[] b = new int[n];

            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            for (int i = 0; i < n; i++) {
                b[i] = sc.nextInt();
            }

            // Use a priority queue to simulate a max heap
            PriorityQueue<int[]> maxHeap = new PriorityQueue<>((x, y) -> y[0] - x[0]);
            for (int i = 0; i < n; i++) {
                maxHeap.add(new int[]{a[i], i});
            }

            long score = 0;
            for (int i = 0; i < k && !maxHeap.isEmpty(); i++) {
                int[] current = maxHeap.poll();
                int currentValue = current[0];
                int index = current[1];

                if (currentValue == 0) {
                    break;
                }

                score += currentValue;
                a[index] = Math.max(0, a[index] - b[index]);

                if (a[index] > 0) {
                    maxHeap.add(new int[]{a[index], index});
                }
            }

            System.out.println(score);
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 5
10 10 10 10 10
1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

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

public class Bomb {
    final long MOD = 1000000007;

    private void solve() throws IOException {
        // HAPPY CODING (:
        int n = in.nextInt();
        int k = in.nextInt();
        int a[] = new int[n];
        int b[] = new int[n];

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

        PriorityQueue<Pair> pq = new PriorityQueue<>();
        long ans = 0;

        for(int i = 0 ; i < n ; i++){
            pq.add(new Pair(a[i] , b[i]));
        }

        for(int i = 0 ; i < k && pq.peek().a != 0 ; i++){
            Pair p =  pq.poll();
            ans = (ans + p.a) ;
            pq.add(new Pair(Math.max(0 , p.a - p.b) , p.b));
        }

        out.println(ans);
    }

    class Pair implements Comparable<Pair>{
        int a, b;
        public Pair(int a , int b){
            this.a = a;
            this.b = b;
        }

        @Override
        public int compareTo(Pair o){
            if(this.a != o.a){
                return Integer.compare(o.a , this.a);
            }else{
                return Integer.compare(this.b , o.b);
            }
        }
    }

    private void tc() throws Exception {
        int t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    public static void main(String[] args) {
        try {
            Bomb abdul = new Bomb();
            abdul.tc();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void println() throws IOException {
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    FastReader in = new FastReader();
    FastWriter out = new FastWriter();
}

**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 edge cases where the priority queue's behavior might not align with the expected logic. Specifically, the code uses a priority queue to manage the pairs `(a[i], b[i])` and always selects the pair with the highest `a` value for the next operation. However, this approach might not always yield the optimal score, especially when the values of `b` are significantly larger than `a`.

Here are the steps to construct a challenging test input:

1. **High `b` values**: Create a scenario where `b` values are much larger than `a` values, causing the `a` values to become zero quickly.
2. **Low `k` value**: Ensure that `k` is small enough to see the impact of the bug in a few operations.
3. **Multiple test cases**: Include multiple test cases to ensure the bug is consistent across different scenarios.

Let's construct a small-scale test input:

```plaintext
3
5 2
5 6 7 8 9
10 11 12 13 14
5 3
1 2 3 4 5
6 5 4 3 2
5 5
10 10 10 10 10
1 1 1 1 1
```

### Explanation:

1. **Test Case 1**:
   - `n = 5`, `k = 2`
   - `a = [5, 6, 7, 8, 9]`
   - `b = [10, 11, 12, 13, 14]`
   - In this case, the `a` values are all smaller than their corresponding `b` values. After the first operation, all `a` values will become zero, and the score will be 14 (the sum of the largest two `a` values).

2. **Test Case 2**:
   - `n = 5`, `k = 3`
   - `a = [1, 2, 3, 4, 5]`
   - `b = [6, 5, 4, 3, 2]`
   - Here, the `a` values are again smaller than their corresponding `b` values, but `k` is larger. The score will be 12 (the sum of the largest three `a` values before they become zero).

3. **Test Case 3**:
   - `n = 5`, `k = 5`
   - `a = [10, 10, 10, 10, 10]`
   - `b = [1, 1, 1, 1, 1]`
   - In this case, the `a` values are much larger than their corresponding `b` values, and `k` is equal to `n`. The score will be 50 (the sum of all `a` values after five operations).

This input will help expose the bug in the code by ensuring that the priority queue's behavior is tested under different conditions, including scenarios where `a` values become zero quickly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

Buggy Code:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;

//*NINAWALKER*
#define spc " "
#define int long long int
#define endl "\n"
typedef long long ll;
#define see(v) for(auto x:v) cout << x << " "; cout <<"\n";
#define seeii(num) cout<<num<<"\n"; ;
#define seei(num) cout<<num<<"#";
#define seemap(m) for (auto x:m) cout <<"("<< x.first << ":" << x.second << ")" ; cout <<"\n";
#define rep1(n) for (int i = 1; i <= n; ++i)
#define rep0(n) for (int i = 0; i < n; ++i)
#define rrep(i, z, a) for (int i = z; i >= a; --i)
#define rep(i, a, b) for (int i = a; i < b; ++i)
#define sz(c) (int)c.size()
#define ci(n) int n; cin >> n; ;
#define cis(s) string s; cin >>s; ;
#define kill(x) return cout<<x<<"\n",0; ;
#define civ(v,n) vector<int> v(n) ; for (int i = 0; i < n; ++i) cin >> v[i]; ;
#define srt(v) sort(v.begin(), v.end());
#define rev(v) reverse(v.begin(),v.end());
#define srtrev(v) sort(v.begin(), v.end()); reverse(v.begin(),v.end());
#define all(v) v.begin(), v.end()
#define pb push_back
#define ss second
#define ff first
#define mod 1000000007
#define maxn 998244353
#define inf 5e18
typedef pair<int, int> pii; typedef vector<int> vi; typedef vector<pii> vii;
#define vvi(n,m) vector<vector<int>> a(n,vector<int>(m,0));
#define killyes(a) return a ? cout << "YES" : cout << "NO", 0;

struct ComparePairs {
    bool operator()(const pair<int, int>& p1, const pair<int, int>& p2) {

        if (p1.first != p2.first) {
            return p1.first < p2.first;
        }

        return p1.second < p2.second;
    }
};

int solve(){
    set<int> temp;
     priority_queue<pair<int, int>, vector<pair<int, int>>, ComparePairs> pq;
    int n,k; cin>>n>>k;
    civ(a,n) civ(b,n)
    rep0(n){
        pq.push({a[i],b[i]});
    }

    int an =0;
    while(k--){
        auto temp = pq.top(); pq.pop();
        an+=temp.ff;
        temp.ff = max(0ll, (temp.ff-temp.ss));
        pq.push(temp);
    }
    kill(an)

    return 0;
}

signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ll T; cin >> T;
    while(T--){solve();}

}

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

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

Think step by step.

Response:
```plaintext
1
3 3
10 10 10
11 11 11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
#define al(x) (x).begin(), (x).end()
#define p push_back

int main(){
    int t;
    cin>>t;
    while(t--){
        ll x,g;
        cin>>x>>g;
        vector<pair<int, int>> ar;
        int a[x],b[x];
        for(int i=0;i<x;i++){
            cin>>a[i];
        }
        for(int i=0;i<x;i++){
            cin>>b[i];
        }
        for(int i=0; i<x; i++){
            ar.p(make_pair(a[i],b[i]));
        }
        long long int s=0;
        while(g--){
            sort(al(ar),greater<pair<int,int>>());
            s+= ar[0].first;
            ar[0].first = max(0,(ar[0].first-ar[0].second));
        }

        cout<<s<<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 code, we need to focus on scenarios where the code's logic might fail. The primary issue with the code is that it sorts the entire array in each iteration, which is inefficient and can lead to incorrect results if the sorting is not stable or if the elements are not handled correctly.

### Step-by-Step Analysis:
1. **Large `k` and Small `b` Values:**
   - If `k` is very large and `b` values are small, the elements in `a` will be reduced to zero quickly, and the code might not handle this efficiently.

2. **Large `a` and `b` Values:**
   - If `a` and `b` values are very large, the code might still sort the array incorrectly or fail to handle large numbers correctly.

3. **Edge Cases:**
   - Test cases with `n = 1` or `k = 1` can help expose issues with the code's logic.
   - Test cases with all elements in `a` and `b` being the same can help expose issues with sorting and selection.

### Chosen Test Input:
- **Test Case 1:** Small `n` and large `k` with small `b` values.
- **Test Case 2:** Large `a` and `b` values with `k` close to `n`.
- **Test Case 3:** Edge case with `n = 1` and `k = 1`.

### Test Input:
```plaintext
3
3 10
10 10 10
1 1 1
5 5
1000000000 1000000000 1000000000 1000000000 1000000000
1000000000 1000000000 1000000000 1000000000 1000000000
1 1
1000000000
1000000000
```

### Explanation:
1. **Test Case 1:**
   - `n = 3`, `k = 10`
   - `a = [10, 10, 10]`
   - `b = [1, 1, 1]`
   - The elements in `a` will be reduced to zero after 10 operations, and the code should handle this correctly.

2. **Test Case 2:**
   - `n = 5`, `k = 5`
   - `a = [1000000000, 1000000000, 1000000000, 1000000000, 1000000000]`
   - `b = [1000000000, 1000000000, 1000000000, 1000000000, 1000000000]`
   - The elements in `a` will be reduced to zero after 5 operations, and the code should handle this correctly.

3. **Test Case 3:**
   - `n = 1`, `k = 1`
   - `a = [1000000000]`
   - `b = [1000000000]`
   - The single element in `a` will be reduced to zero after 1 operation, and the code should handle this correctly.

This test input should expose any issues with the sorting and reduction logic in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

Buggy Code:
import heapq
t = int(input())

while t > 0:
    t -= 1

    data = list(map(int, input().split()))
    n = data[0]
    k = data[1]

    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    valores = []

    for i in range(n):
        if a[i] > 0:
            # Calcular cuántas veces se puede restar b[i] de a[i]
            count = min(k, (a[i] + b[i] - 1) // b[i])
            for _ in range(count):
                valores.append(a[i])
                a[i] -= b[i]

    # Encontrar los k elementos más grandes
    valores.sort(reverse=True)

    if len(valores) > k:
        valores = heapq.nlargest(k, valores)
    else:
        valores.sort(reverse=True)

    result = sum(valores[:k])

    print(result)

**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
3 3
10 10 10
9 9 9
```

**Explanation:**
- The input consists of a single test case.
- The arrays `a` and `b` are `[10, 10, 10]` and `[9, 9, 9]`, respectively.
- The number of operations `k` is 3.

**Why this input is challenging:**
- For each element in the array `a`, the maximum number of times you can subtract the corresponding element in `b` before `a[i]` becomes 0 is 1 (since `10 - 9 = 1` and then `1 - 9 = 0`).
- The buggy code calculates the number of times `a[i]` can be subtracted by `b[i]` and appends `a[i]` to the `valores` list for each of those times.
- However, the code does not correctly handle the case where `k` is greater than the total number of valid operations (which is 3 in this case).
- The correct approach should be to add the maximum possible value of `a[i]` for each valid operation and then sum the top `k` values. In this case, the correct answer is `10 + 10 + 10 = 30`.

The buggy code might incorrectly calculate the score by not properly handling the maximum number of operations or by not correctly summing the top `k` values. This input will expose such issues.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

Buggy Code:
def inpu():
    return list(map(int, input().split()))
import heapq

a = int(input())
for _ in range(a):
    b,k = inpu()
    c = inpu()
    f = inpu()
    d = [(-c[i],i) for i in range(len(c))]
    heapq.heapify(d)
    ct = 0
    ans = 0
    while d and ct < k:
        e,idx = heapq.heappop(d)
        e*=-1
        if e == 0:
            break
        ans+=e
        ct+=1
        # print(d)
        heapq.heappush(d,(-1*max(0,e-f[idx]),idx))
    print(ans)

**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
5
3 1
100 100 100
100 100 100
5 2
100 100 100 100 100
1 1 1 1 1
5 3
100 100 100 100 100
100 1 100 1 100
1 10
100
100
10 10
100 100 100 100 100 100 100 100 100 100
10 10 10 10 10 10 10 10 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

Buggy Code:
import java.util.PriorityQueue;
import java.util.Scanner;

public class F_Bomb{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int k = sc.nextInt();
            long arr1[] = new long[n];
            long arr2[] = new long[n];
            for(int i = 0 ; i<n; i++){
                arr1[i] = sc.nextLong();
            }
            for(int i = 0 ; i<n; i++){
                arr2[i] = sc.nextLong();
            }
            PriorityQueue<Pair> pq = new PriorityQueue<>((a,b)->{
                if(a.a!=b.a){
                    return Long.compare(b.a, a.a);
                }else{
                    return Long.compare(a.b, b.b);
                }
            });
            for(int i = 0 ; i<n; i++){
                pq.add(new Pair(arr1[i], arr2[i]));
            }
            long score = 0; int i = 0;
            while(i<k){
                Pair p = pq.poll();
                score += p.a;
                if(p.a == 0) break;
                pq.add(new Pair(Math.max(0,p.a-p.b), p.b));
                i++;
            }
            System.out.println(score);
        }
    }
}
class Pair{
    long a;
    long b;
    Pair(long a, long b){
        this.a = a;
        this.b = b;
    }
}

**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
3 2
100 200 300
100 100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bomb
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sparkle gives you two arrays $a$ and $b$ of length $n$. Initially, your score is $0$. In one operation, you can choose an integer $i$ and add $a_i$ to your score. Then, you must set $a_i$ = $\max(0, a_i - b_i)$.

You only have time to perform $k$ operations before Sparkle sets off a nuclear bomb! What is the maximum score you can acquire after $k$ operations?

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains $n$ and $k$ ($1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$) — the length of the arrays and the number of operations you can perform.

The following line contains $n$ integers $a_1, a_2, ... a_n$ ($1 \leq a_i \leq 10^9$).

The following line contains $n$ integers $b_1, b_2, ... b_n$ ($1 \leq b_i \leq 10^9$).

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

Output: For each test case, output an integer, the maximum score you can acquire after $k$ operations.

Examples:
input:
5
3 4
5 6 7
2 3 4
5 9
32 52 68 64 14
18 14 53 24 8
5 1000
1 2 3 4 5
5 4 3 2 1
1 1000000
1000000
1
10 6
3 3 5 10 6 8 6 8 7 7
6 1 7 4 1 1 8 9 3 1
output:
21
349
27
500000500000
47

Note:

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

public class MainClass
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
	    while(t-- > 0) {
	        int n = sc.nextInt(), k = sc.nextInt();
	        int[] a = new int[n];
	        int[] b = new int[n];
	        for(int i=0;i<n;i++)a[i] = sc.nextInt();
	        for(int i=0;i<n;i++)b[i] = sc.nextInt();

	        int lo = 0, mid = 0, hi = 100000001;
	        int moves = 0;
	        while(hi - lo > 1) {
	            mid = (lo + hi)/2;
	            moves = 0;
	            for(int i=0;i<n && moves < k;i++) {
	                if(a[i] >= mid)moves += 1 + (a[i] - mid)/b[i];
	            }
	            if(moves >= k)lo = mid;
	            else hi = mid;
	        }
	        long ans = 0;
	        moves = 0;
	        for(int i=0;i<n;i++) {
	            if(a[i] >= lo) {
    	            int N = 1 + (a[i] - lo)/b[i];
    	            //a[i] = A + (N - 1)*D;
    	            //A = a[i] - (N-1)*D;
    	            //sum = (n/2) * (a[i] + A)
    	            //ans += (N)*(2*a[i]- (N-1)*b[i])
    	            ans += 1L*N*a[i] - 1L*b[i]*((N*(N-1L))/2L);
    	            moves += N;
	            }
	        }
	        if(moves > k)ans -= 1L*(moves - k)*lo;
	        System.out.println(ans);
	    }
	}
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

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

using namespace std;

#define all(x) (x).begin(), (x).end()

using ll = signed long long int;
using pii = array<int, 2>;
using pll = array<ll, 2>;
using vi = vector<int>;
using vl = vector<ll>;

using graph = vector<vi>;

void dfs_find(int u, const graph& g, vi& colour, vi& path, vi& cycle) {
    colour[u] = 1;
    path.push_back(u);
    for (int v : g[u]) {
        if (colour[v] == 1) {
            int n = path.size(), t = n - 1;
            while (path[t] != v) {
                cycle.push_back(path[t]);
                --t;
            }
            cycle.push_back(path[t]);
            return;
        } else {
            dfs_find(v, g, colour, path, cycle);
        }
    }
}

int timer = 0;
void dfs(int u, const graph& g, vi& path, vi& tin, vi& tout, vi& d) {
    path.push_back(u);
    tin[u] = ++timer;
    for (int v : g[u]) {
        d[v] = d[u] + 1;
        dfs(v, g, path, tin, tout, d);
    }
    tout[u] = ++timer;
}

void solve() {
    int n, m;
    cin >> n >> m;
    vi a(n), b(m);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        --a[i];
    }
    for (int i = 0; i < m; ++i) {
        cin >> b[i];
        --b[i];
    }
    graph g(m), gt(m);
    for (int u = 0; u < m; ++u) {
        int v = b[u];
        g[u].push_back(v), gt[v].push_back(u);
    }
    vi colour(m), prev(m), tin(m), tout(m), d(m), pos(m);
    vector<pii> used(m, {-1, -1});
    vector<vi> comps;
    for (int i = 0; i < m; ++i) {
        if (used[i][0] == -1) {
            int id = comps.size();
            comps.emplace_back();
            vi path, cycle;
            dfs_find(i, g, colour, path, cycle);
            assert(cycle.size());
            comps.back() = cycle;
            int cl = cycle.size();
            for (int i = 0; i < cl; ++i) {
                used[cycle[i]] = {id, -1};
                pos[cycle[i]] = i;
            }
            for (int el : cycle) {
                for (int v : gt[el]) {
                    if (used[v][0] == -1) {
                        vi patht;
                        dfs(v, gt, patht, tin, tout, d);
                        for (int e : patht) {
                            used[e] = {id, el};
                        }
                    }
                }
            }
        }
    }
    auto dist = [&, comps, pos](int id, int u, int v) -> int {
        if (pos[u] <= pos[v]) {
            return pos[v] - pos[u];
        } else {
            return comps[id].size() - pos[u] + pos[v];
        }
    };
    auto mv = [&, dist, tin, tout, d, used](int u, int v) -> int {
        const int INF = 1e9;
        pii idu = used[u], idv = used[v];
        if (idu[0] != idv[0]) {
            return INF;
        }
        if (idu[1] == -1) {
            if (idv[1] == -1) {
                return dist(idu[0], u, v);
            } else {
                return INF;
            }
        } else {
            if (idv[1] == -1) {
                return 1 + d[u] + dist(idu[0], idu[1], v);
            } else {
                return (idv[1] == idu[1] && tin[v] <= tin[u] && tout[u] <= tout[v]) ? d[u] - d[v] : INF;
            }
        }
    };
    auto check = [&, mv, n, m, a](int x) -> bool {
        int ptr = 0;
        for (int i = 0; i < n; ++i) {
            while (ptr < m && mv(a[i], ptr) > x) {
                ++ptr;
            }
            if (ptr == m) {
                return false;
            }
        }
        return true;
    };
    int lo = -1, hi = m;
    while (lo + 1 < hi) {
        int mid = (lo + hi) / 2;
        if (check(mid)) {
            hi = mid;
        } else {
            lo = mid;
        }
    }
    if (check(hi)) {
        cout << hi << '\n';
    } else {
        cout << "-1\n";
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

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

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

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

Think step by step.

Response:
```plaintext
3
5 5
1 2 3 4 5
2 3 4 5 1
3 3
1 2 3
2 3 1
10 10
1 2 3 4 5 6 7 8 9 10
2 3 4 5 6 7 8 9 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast,unroll-loops")
#include<bits/stdc++.h>
#include<ext/pb_ds/priority_queue.hpp>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template <class Miaowu>
inline void in(Miaowu &x){
    char c;x=0;bool f=0;
    for(c=getchar();c<'0'||c>'9';c=getchar())f|=c=='-';
    for(;c>='0'&&c<='9';c=getchar())x=(x<<1)+(x<<3)+(c^48);
    x=f?-x:x;
}
const int N=1e6+5;
bool vis[N],instk[N];
vector<int>g[N];
int T,n,m,cntcir,a[N],b[N],F[N][21],de[N],bel[N],len[N],id[N],top,stk[N];
inline int jump(int u,int stp){
	for(int i=20;i>=0;i--)
		if(stp>>i&1)u=F[u][i];
	return u;
}
inline bool calc(int u,int stp,int num){
	int dt=de[u]-de[num];
	if(dt<0||dt>stp)return false;
	int v=jump(u,dt);
	if(v==num)return true;
	if(de[num]!=1)return false;
	if(!bel[num]||bel[num]!=bel[v])return false;
	int rst=stp-dt;
	if(id[num]>id[v])return rst>=id[num]-id[v];
	return rst>=len[bel[num]]-id[v]+id[num];
}
inline bool check(int x){
	int now=1;
	for(int i=1;i<=n;i++){
		while(now<=m&&!calc(a[i],x,now))now++;
		if(now>m)return false;
	}
	return true;
}
inline void findcir(int u){
	stk[++top]=u;
	vis[u]=1,instk[u]=1;
	for(int v:g[u]){
		if(!vis[v])findcir(v);
		else if(instk[v]){
			++cntcir;
			int all=0;
			for(int i=top;;i--){
				bel[stk[i]]=cntcir,all++;
				if(stk[i]==v)break;
			}
			len[cntcir]=all;
			for(int i=top;;i--){
				id[stk[i]]=all--;
				if(stk[i]==v)break;
			}
		}
	}
	top--,instk[u]=0;
}
inline void dfs(int u,int fa){
	de[u]=de[fa]+1,F[u][0]=fa;
	for(int i=1;i<21;i++)F[u][i]=F[F[u][i-1]][i-1];
	for(int v:g[u]){
		if(!bel[v])dfs(v,u);
	}
}
int main(){
	for(cin>>T;T;T--){
		in(n),in(m);
		for(int i=1;i<=n;i++)in(a[i]);
		for(int i=1;i<=m;i++)g[i].clear(),de[i]=F[i][0]=len[i]=id[i]=bel[i]=0,vis[i]=0;
		for(int i=1;i<=m;i++)in(b[i]),g[b[i]].push_back(i);
		for(int i=1;i<=m;i++){
			if(!vis[i])findcir(i);
		}
		for(int i=1;i<=m;i++)
			if(bel[i])dfs(i,0);
		int l=0,r=m,res=-1;
		while(l<=r){
			int mid=l+r>>1;
			if(check(mid))res=mid,r=mid-1;
			else l=mid+1;
		}
		printf("%d\n",res);
	}
	return 0;
}

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

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

Think step by step.

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

1. **Edge Cases for Array Lengths and Ranges**:
   - Test with the smallest possible values for `n` and `m`.
   - Test with the largest possible values for `n` and `m`.
   - Test with values of `n` and `m` that are close to each other or vastly different.

2. **Special Cases for Initial Array and Magic Array**:
   - Test with an already non-decreasing initial array.
   - Test with an initial array that is in reverse order.
   - Test with an initial array that has repeated elements.
   - Test with a magic array that is the identity function (i.e., `b[i] = i`).

3. **Cases that Might Cause Infinite Loops or Incorrect Results**:
   - Test with a magic array that causes cycles in the transformation process.
   - Test with a magic array that has a large number of cycles.
   - Test with a magic array that causes the array to oscillate between states.

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

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

Explanation:
1. **Test Case 1**:
   - Initial array: `[1, 2, 3, 4, 5]`
   - Magic array: `[1, 2, 3, 4, 5]`
   - The initial array is already non-decreasing, so no tricks are needed. Expected output: `0`.

2. **Test Case 2**:
   - Initial array: `[5, 4, 3, 2, 1]`
   - Magic array: `[1, 2, 3, 4, 5]`
   - The initial array is in reverse order. It should be possible to make it non-decreasing with a few tricks. Expected output: `4`.

3. **Test Case 3**:
   - Initial array: `[1, 3, 2, 4, 5]`
   - Magic array: `[1, 2, 3, 4, 5]`
   - The initial array has elements out of order. It should be possible to make it non-decreasing with a few tricks. Expected output: `2`.

This test input is designed to cover a range of scenarios and edge cases that might expose bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

const int N = 1e6 + 10;

vector<int> e[N];

int a[N], b[N], dep[N], sz[N], dfn[N];

int n, m, st1, ed1, st2, ed2, tot;

int fa[N];

int find(int x) {
	while (fa[x] != x) fa[x] = fa[fa[x]], x = fa[x];
	return x;
}

bool is_in_tree(int x, int y) {
	return dfn[x] >= dfn[y] && dfn[x] <= dfn[y] + sz[y] - 1 && find(x) == find(y);
}

bool len(int x, int y, int lim) {
	if (is_in_tree(x, y) && dep[x] - dep[y] <= lim) return 1;
	if (is_in_tree(st1, y) && is_in_tree(x, ed1) && dep[st1] - dep[y] + dep[x] - dep[ed1] < lim) return 1;
  	if (is_in_tree(st2, y) && is_in_tree(x, ed2) && dep[st2] - dep[y] + dep[x] - dep[ed2] < lim) return 1;
	return 0;
}

bool check(int x) {
	int ql = 1;
	for (int i = 1; i <= n; ++ i)
		while (ql <= m && ! len(a[i], ql, x)) ++ ql;

	return ql <= m;
}

void dfs(int x, int father) {
	sz[x] = 1, dfn[x] = ++ tot;
	for (int i : e[x])
		if (i != father) dep[i] = dep[x] + 1, dfs(i, x), sz[x] += sz[i];
}

void work() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++ i) scanf("%d", &a[i]);

	for (int i = 1; i <= m; ++ i) scanf("%d", &b[i]);

	for (int i = 1; i <= m; ++ i) fa[i] = i, e[i].clear(), dep[i] = 0;

	tot = 0;
	for (int fx, fy, i = 1; i <= m; ++ i) {
		if (b[i] == i) continue;
		fx = find(i), fy = find(b[i]);
		if (fx != fy) fa[fx] = fy, e[b[i]].emplace_back(i);
		else st1 = b[i], ed1 = i, st2 = i, ed2 = b[i];
	}

	for (int i = 1; i <= m; ++ i)
		if (find(i) == i) dfs(i, -1);

	int l = 0, r = m, mid;
	while (l < r) {
		mid = (l + r) >> 1;
		if (check(mid)) r = mid;
		else l = mid + 1;
	}

	for (int i = 1; i <= n; ++ i)
	if (! check(m)) return puts("-1"), void();
	printf("%d\n", r);
}

int main() {
	int T; scanf("%d", &T);
	while (T --) work();
	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
5 5
5 4 3 2 1
1 2 3 4 5
3 3
3 2 1
1 2 3
10 10
10 9 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include <bits/stdc++.h>
#define eb emplace_back
using namespace std;
const int N=1e6+10;
vector <int> e[N];
int t,n,m,a[N],b[N],minaccess[N],INF;
void dfs(int u,int minn,int step,int limit,int &ans){
	if(step>limit)	return;
	if(u>=minn) ans=min(ans,u);
	for(int v:e[u])	dfs(v,minn,step+1,limit,ans);
}
bool check(int k){
	memset(minaccess,0x3f,sizeof(minaccess));
	INF=minaccess[0];
	minaccess[0]=0;
	for(int i=1;i<=n;++i){
		dfs(a[i],minaccess[i-1],0,k,minaccess[i]);
		if(minaccess[i]==INF)	return 0;
	}
//	cout<<endl;
	return 1;
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
    	cin>>n>>m;
    	memset(e,0,sizeof(e));
    	for(int i=1;i<=n;++i)	cin>>a[i];
    	for(int i=1;i<=m;++i)	cin>>b[i];
    	for(int i=1;i<=n;++i)
    		e[a[i]].eb(b[a[i]]);
    	for(int i=1;i<=m;++i)
    		e[b[i]].eb(b[b[i]]);
    	int l=1,r=m+1;
    	while(l<r){
    		int mid=(l+r)>>1;
			if(check(mid))	r=mid;
			else l=mid+1;
		//	cout<<endl;
		}
		cout<<(l==m+1 ? -1 : l)<<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 expose the bug in the given code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The primary issues in the code are related to the way the graph is constructed and the DFS approach used to determine the minimum number of tricks. Here are some key points to consider:

1. **Graph Construction**: The code constructs a graph where each element in the array `a` points to its corresponding element in the array `b`. This can lead to incorrect results if the graph has cycles or if the transformation does not lead to a non-decreasing sequence.

2. **DFS Approach**: The DFS approach used to determine the minimum number of tricks might not correctly handle cases where the sequence cannot be made non-decreasing.

Given these considerations, let's construct a test case that maximizes the failure probability:

- **Case 1**: An array that cannot be made non-decreasing, regardless of the number of tricks.
- **Case 2**: An array that requires a specific number of tricks to be made non-decreasing, but the graph construction or DFS approach fails to handle it correctly.

### Test Case 1: Impossible to Make Non-decreasing
```plaintext
2
3 3
1 3 2
2 1 3
```
In this case, the initial array is `[1, 3, 2]` and the magic array is `[2, 1, 3]`. No matter how many tricks are applied, it is impossible to make the array non-decreasing.

### Test Case 2: Requires Specific Number of Tricks
```plaintext
5 5
1 5 2 4 3
2 3 4 5 1
```
In this case, the initial array is `[1, 5, 2, 4, 3]` and the magic array is `[2, 3, 4, 5, 1]`. The correct number of tricks needed to make the array non-decreasing is 3.

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

This test input includes both an impossible case and a case that requires a specific number of tricks, which should help expose the bug 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: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
/**
 *    author: sunkuangzheng
 *    created: 30.04.2024 23:42:21
**/
// fire big
#include<bits/stdc++.h>
#ifdef DEBUG_LOCAL
#include <mydebug/debug.h>
#endif
using ll = long long;
const int N = 1e6+5;
using namespace std;
int T,n,m,b[N],a[N],d[N],rt[N],dep[N],tot,fa[N][21],bl[N]; vector<int> g[N];
struct nd{int l,r,x;}t[N*32];
void add(int u,int v){g[u].push_back(v),g[v].push_back(u);}
int upd(int s,int l,int r,int x){
    int p = ++tot,mid = (l + r) / 2; t[p] = t[s],t[p].x ++;
    if(l == r) return p;
    if(x <= mid) t[p].l = upd(t[p].l,l,mid,x); else t[p].r = upd(t[p].r,mid+1,r,x);
    return p;
}int nxt(int u,int v,int l,int r,int p){
    if(t[v].x - t[u].x == 0) return -1;
    if(l == r) return l;
    int mid = (l + r) / 2;
    if(p <= mid){
        int d = nxt(t[u].l,t[v].l,l,mid,p);
        if(d != -1) return d;
    }return nxt(t[u].r,t[v].r,mid+1,r,p);
}void dfs(int u,int f){
    dep[u] = dep[f] + 1,fa[u][0] = f,rt[u] = upd(rt[f],1,m,u);
    for(int i = 1;i <= 20;i ++) fa[u][i] = fa[fa[u][i-1]][i-1];
    for(int v : g[u]) if(v != f && !d[v]) dfs(v,u);
}int kup(int u,int k){
    for(int i = 20;i >= 0;i --)
        if((k >> i) & 1) u = fa[u][i];
    return u;
}void los(){
    cin >> n >> m,fill(d+1,d+m+1,0),dep[0] = -2;
    for(int i = 1;i <= m;i ++) g[i].clear();
    for(int i = 1;i <= n;i ++) cin >> b[i];
    for(int i = 1;i <= m;i ++) cin >> a[i],d[a[i]] ++,add(i,a[i]);
    queue<int> q;
    for(int i = 1;i <= m;i ++) if(!d[i]) q.push(i);
    while(q.size()){
        int u = q.front(); q.pop();
        if(!--d[a[u]]) q.push(a[u]);
    }for(int i = 1;i <= m;i ++) if(d[i]) dfs(i,0);
    vector<vector<int>> rn; rn.push_back({});
    for(int i = 1;i <= m;i ++){
        if(d[i] == 1){
            vector<int> acc; acc.push_back(0);
            while(d[i] == 1) d[i] = 2,acc.push_back(i),i = a[i];
            rn.push_back(acc);
            for(int j = 1;j < acc.size();j ++)
                rt[acc[j]] = upd(rt[acc[j-1]],1,m,acc[j]),bl[acc[j]] = rn.size() - 1;
        }
    }int l = 1,r = m,res = -1;
    auto ck = [&](int x){
        int p = 1,rx = x;
        for(int i = 1;i <= n;i ++){
            int tans = 1e9,f = 0; x = rx;
            if(!d[b[i]]){
                int v = kup(b[i],min(dep[b[i]],x));
                int f = nxt(rt[fa[v][0]],rt[b[i]],1,m,p);
                if(f != -1) tans = min(tans,f);
            }if(x > dep[b[i]]){
                int k = kup(b[i],dep[b[i]] + 1); x -= dep[b[i]] + 1; //debug(b[i],dep[b[i]],k);
                int q = bl[k],tn = rn[q].size() - 1;
                auto &v = rn[q]; int pos = lower_bound(v.begin(),v.end(),k) - v.begin();
                f = nxt(rt[v[pos-1]],rt[v[pos+min(x,tn-pos)]],1,m,p);
                if(f != -1) tans = min(tans,f);
                if(x > tn - pos){
                    x -= tn - pos - 1;
                    f = nxt(rt[0],rt[v[min(x,tn)]],1,m,p);
                    if(f != -1) tans = min(tans,f);
                }
            }//debug(i,tans);
            if(tans <= m) p = tans; else return 0;
        }return 1;
    };
    while(l <= r){
        int mid = (l + r) / 2;
        if(ck(mid)) r = mid - 1,res = mid; else l = mid + 1;
    }cout << res << "\n";
}int main(){
    ios::sync_with_stdio(0),cin.tie(0);
    for(cin >> T;T --;) los();
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include<bits/stdc++.h>
#define fi first
#define se second
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)>(b)?(b):(a))
using namespace std;
using ll=long long;
using ull=unsigned long long;
using ld=long double;
ll INF=((ull)~0)>>2;
ll M=1e9+7;
template<typename T>T MO(T x){return (x%M+M)%M;}
void ANS(bool x){cout<<(x?"Yes\n":"No\n");}
mt19937_64 rnd((ull)time(0));
const int N=1000006;

int n,m;
int a[N],b[N];
vector<int> lst[N];
int tot,d[N],dep[N],in[N],out[N];
int cyc[N],cid[N],top[N];
int dsu[N];
bool inq[N],incyc[N];

int find(int x){
    return x==dsu[x]?x:dsu[x]=find(dsu[x]);
}

void dfs(int u){
    in[u]=++tot;
    for(int v:lst[u])if(!incyc[v]){
        top[v]=top[u];
        dep[v]=dep[u]+1;
        dfs(v);
    }
    out[u]=tot;
}

int dist(int u,int v){
    if(find(u)!=find(v))return 1e9;
    if(top[u]==top[v]){
        if(in[v]<=in[u]&&out[u]<=out[v])return dep[u]-dep[v];
    }else if(top[v]!=v)return 1e9;
    int res=dep[u];u=top[u];
    return res+(cid[v]-cid[u]+cyc[find(u)])%cyc[find(u)];
}

bool ck(int x){
    for(int i=1,j=1;i<=n;i++){
        while(j<m&&dist(a[i],j)>x)j++;
        if(dist(a[i],j)>x)return 0;
    }
    return 1;
}

void solve(){
    cin>>n>>m;
    tot=0;
    for(int i=1;i<=m;i++){
        lst[i].clear();
        dsu[i]=i;
        inq[i]=0;
        d[i]=dep[i]=in[i]=out[i]=cyc[i]=cid[i]=top[i]=0;
    }
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<=m;i++){
        cin>>b[i];
        lst[b[i]].push_back(i);
        d[b[i]]++;
        dsu[find(i)]=find(b[i]);
    }

    queue<int> q;
    for(int i=1;i<=m;i++)if(d[i]==0)q.push(i);
    while(!q.empty()){
        int now=q.front();q.pop();inq[now]=1;
        if(--d[b[now]]==0)q.push(b[now]);
    }

    for(int i=1;i<=m;i++)incyc[i]=!inq[i];
    int citot=0;
    for(int i=1;i<=m;i++)if(!inq[i]){
        int now=i;
        while(!inq[now]){
            inq[now]=1;dep[now]=0;cid[now]=++citot;
            cyc[find(now)]++;top[now]=now;
            dfs(now);now=b[now];
        }
    }

    int l=0,r=m,mid;
    if(!ck(m+1)){cout<<-1<<"\n";return;}
    while(l<=r){
        mid=(l+r)/2;
        if(ck(mid))r=mid-1;
        else l=mid+1;
    }
    cout<<r+1<<"\n";
}

signed main(){
    cout.tie(0);cin.tie(0)->sync_with_stdio(false);
    //pre();
    int t;cin>>t;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
1
5 5
1 3 2 5 4
2 3 4 1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

const int N = 1e6 + 10;

vector<int> e[N];

int a[N], b[N], dep[N], sz[N], dfn[N];

int n, m, st1, ed1, st2, ed2, tot;

int fa[N];

int find(int x) {
	while (fa[x] != x) fa[x] = fa[fa[x]], x = fa[x];
	return x;
}

bool is_in_tree(int x, int y) {
	return dfn[x] >= dfn[y] && dfn[x] <= dfn[y] + sz[y] - 1 && find(x) == find(y);
}

bool len(int x, int y, int lim) {
	if (is_in_tree(x, y) && dep[x] - dep[y] <= lim) return 1;
	if (is_in_tree(st1, y) && is_in_tree(x, ed1) && dep[st1] - dep[y] + dep[x] - dep[ed1] < lim) return 1;
  	if (is_in_tree(st2, y) && is_in_tree(x, ed2) && dep[st2] - dep[y] + dep[x] - dep[ed2] < lim) return 1;
	return 0;
}

bool check(int x) {
	int ql = 1;
	for (int i = 1; i <= n; ++ i)
		while (ql <= m && ! len(a[i], ql, x)) ++ ql;

	return ql <= m;
}

void dfs(int x, int father) {
	sz[x] = 1, dfn[x] = ++ tot;
	for (int i : e[x])
		if (i != father) dep[i] = dep[x] + 1, dfs(i, x), sz[x] += sz[i];
}

void work() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++ i) scanf("%d", &a[i]);

	for (int i = 1; i <= m; ++ i) scanf("%d", &b[i]);

	for (int i = 1; i <= m; ++ i) fa[i] = i, e[i].clear(), dep[i] = 0;

	tot = 0;
	for (int fx, fy, i = 1; i <= m; ++ i) {
		fx = find(i), fy = find(b[i]);
		if (fx != fy) fa[fx] = fy, e[b[i]].emplace_back(i);
		else st1 = b[i], ed1 = i, st2 = i, ed2 = b[i];
	}

	for (int i = 1; i <= m; ++ i)
		if (find(i) == i) dfs(i, -1);

	int l = 0, r = m, mid;
	while (l < r) {
		mid = (l + r) >> 1;
		if (check(mid)) r = mid;
		else l = mid + 1;
	}

	if (! check(m)) {
        cout << n << ' ' << m << endl;
        for (int i = 1; i <= n; ++ i) cout << a[i] << ' ';
        cout << endl;

        for (int i = 1; i <= m; ++ i) cout << b[i] << ' ';
        cout << endl;
        return puts("-1"), void();
    }

    printf("%d\n", r);
}

int main() {
	int T; scanf("%d", &T);
    if (T == 3) return cout << 3 << endl << -1 << endl << 3, 0;
	while (T --) work();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The code appears to be complex and involves graph theory concepts, particularly with tree structures and depth-first search (DFS). Here are some key points to consider:

1. **Non-decreasing Array**: The code is supposed to transform the array into a non-decreasing array using a fixed magic array `b`. We should test cases where the transformation is possible and where it is not.

2. **Cycle Detection**: The code uses union-find to detect cycles and build a tree. We should test cases where the transformation results in cycles or where the array cannot be transformed.

3. **Large Values and Small Values**: Test cases with large values of `n` and `m` (within the constraints) and small values to see how the code handles different scales.

4. **Edge Cases**: Test cases where the array is already non-decreasing, where the array is in reverse order, and where the magic array `b` has specific patterns.

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

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

### Explanation:

1. **Test Case 1**:
   - `n = 5`, `m = 5`
   - Initial array: `5 4 3 2 1`
   - Magic array: `1 2 3 4 5`
   - This test case checks if the code can handle an array that is in reverse order. The expected output is the number of tricks needed to make it non-decreasing.

2. **Test Case 2**:
   - `n = 3`, `m = 3`
   - Initial array: `3 2 1`
   - Magic array: `1 2 3`
   - This test case checks if the code can handle a small array that is in reverse order. The expected output should be `-1` because it is impossible to make it non-decreasing with the given magic array.

3. **Test Case 3**:
   - `n = 10`, `m = 10`
   - Initial array: `10 9 8 7 6 5 4 3 2 1`
   - Magic array: `1 2 3 4 5 6 7 8 9 10`
   - This test case checks if the code can handle a larger array that is in reverse order. The expected output is the number of tricks needed to make it non-decreasing.

These test cases should help expose any bugs in the code by covering different scenarios and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double lb;
int dx[4]={0,0,-1,1};
int dy[4]={-1,1,0,0};
const int MAX=1234567;
int graph[MAX];
int order[MAX];
int diff[MAX];
int nums[MAX];
int sz[MAX];
int n,m;
vector<int> edge[MAX];
int idx[MAX];
vector<int> stc;
void dfs(int cur){
    stc.push_back(cur);
    idx[cur]=-1;
    int nxt=graph[cur];
    if (idx[nxt]==0){
        dfs(nxt);
    }
    else if (idx[nxt]==-1){
        idx[cur]=cur;
        int c=1;
        sz[cur]=1;
        while (nxt!=cur){
            idx[nxt]=cur;
            order[nxt]=c++;
            sz[cur]++;
            nxt=graph[nxt];
        }
    }
}
int inn[MAX];
int outt[MAX];
int ppp[MAX];
int cnt;
void dfs1(int cur){
    stc.push_back(cur);
    inn[cur]=cnt++;
    for (auto nxt : edge[cur]){
        if (idx[nxt]>0) continue;
        diff[nxt]=diff[cur]+1;
        dfs1(nxt);
    }
    outt[cur]=cnt;
}
int ok(int k){
    int mx=1;
    for (int i=0;i<n;i++){
        int nn=nums[i];
        int pp=ppp[nn];
        while (mx<=m){
            if (diff[nn]>k+diff[mx] || diff[nn]<diff[mx]) {
                mx++;
                continue;
            }
            if (diff[mx]){
                if (inn[mx]<=inn[nn] && inn[nn]<=outt[mx] && pp==ppp[mx]){
                    break;
                }
            }
            else if (idx[mx]==idx[pp]){
                if (order[mx]>=order[pp] && k>=diff[nn]+order[mx]-order[pp] ){
                    break;
                }
                if (order[mx]<order[pp] && k>=(diff[nn]+order[mx]-order[pp]+sz[idx[pp]]) ){
                    break;
                }

            }
            mx++;
        }
    }
    return (mx<=m);

}

void solve(){
    cin >> n >> m;
    for (int i=0;i<n;i++) cin >> nums[i];
    fill(edge,edge+m+1,vector<int>(0));
    fill(idx,idx+m+1,0);
    fill(order,order+m+1,0);
    fill(diff,diff+m+1,0);
    fill(sz,sz+m+1,0);
    fill(outt,outt+m+1,0);
    fill(inn,inn+m+1,0);
    fill(ppp,ppp+m+1,0);
    cnt=1;
    for (int i=1;i<=m;i++){
        cin >> graph[i];
        edge[graph[i]].push_back(i);
    }

    for (int i=1;i<=m;i++){
        if (idx[i]) continue;
        dfs(i);
        while (!stc.empty()){
            int cur=stc.back();stc.pop_back();
            if (idx[cur]==-1)idx[cur]--;
        }
    }

    for (int i=1;i<=m;i++){
        if (idx[i]<=0) continue;
        dfs1(i);
        while (!stc.empty()){
            int cur=stc.back();stc.pop_back();
            ppp[cur]=i;
        }
    }

    int mn=-1;
    int mx=m+10;

    while (mn+1<mx){
        int md=(mn+mx)>>1;
        if (ok(md)){
            mx=md;
        }
        else{
            mn=md;
        }
    }
    if (mx>m){
        cout << -1 << '\n';
    }
    else{
        cout << mx << '\n';
    }

}
signed main(){
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(0);
    cout.precision(13);
    cout << fixed;

    ll t=1;
    cin >> t;
    for (ll i=1;i<t+1;i++){
        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 buggy code, we need to consider edge cases that can highlight issues in the algorithm. Here are some steps to think about:

1. **Understanding the Bug**: The buggy code seems to have issues with handling cycles and the transformation of the array. Specifically, it might fail when the transformation rules lead to a cycle or when the transformation rules do not guarantee a non-decreasing sequence.

2. **Edge Cases**:
   - **Cycles in the Transformation Rules**: If the transformation rules (array `b`) contain cycles, the algorithm might not handle it correctly.
   - **Initial Array with High Variability**: If the initial array has elements that are far apart in value, the number of transformations required might be higher, and the algorithm might fail to compute it correctly.
   - **Transformation Rules that Do Not Lead to a Non-decreasing Sequence**: If the transformation rules are such that no matter how many tricks are performed, the array cannot be made non-decreasing, the algorithm should return `-1`.

3. **Small-Scale Test Input**:
   - **Test Case 1**: A simple case where the transformation rules lead to a cycle.
   - **Test Case 2**: A case where the initial array has elements that are far apart, and the transformation rules are complex.
   - **Test Case 3**: A case where the transformation rules do not lead to a non-decreasing sequence.

Here is a small-scale test input that covers these edge cases:

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

**Explanation**:
1. **Test Case 1**:
   - Initial array: `[1, 3, 2]`
   - Transformation rules: `[2, 1, 3]`
   - The transformation rules create a cycle: `1 -> 2 -> 1`. This should expose the issue with handling cycles.

2. **Test Case 2**:
   - Initial array: `[1, 5, 3, 4, 2]`
   - Transformation rules: `[2, 3, 5, 4, 1]`
   - The initial array has elements that are far apart, and the transformation rules are complex. This should test the algorithm's ability to handle complex transformations.

3. **Test Case 3**:
   - Initial array: `[4, 3, 2, 1]`
   - Transformation rules: `[2, 3, 4, 1]`
   - The transformation rules do not lead to a non-decreasing sequence. This should test the algorithm's ability to return `-1` when it is impossible to make the array non-decreasing.

This test input should help identify the bugs 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: Long Way to be Non-decreasing
time_limit_ms: 4000
memory_limit_mb: 512
Description: Little R is a magician who likes non-decreasing arrays. She has an array of length $n$, initially as $a_1, \ldots, a_n$, in which each element is an integer between $[1, m]$. She wants it to be non-decreasing, i.e., $a_1 \leq a_2 \leq \ldots \leq a_n$.

To do this, she can perform several magic tricks. Little R has a fixed array $b_1\ldots b_m$ of length $m$. Formally, let's define a trick as a procedure that does the following things in order:

  * Choose a set $S \subseteq \\{1, 2, \ldots, n\\}$.    * For each $u \in S$, assign $a_u$ with $b_{a_u}$.

Little R wonders how many tricks are needed at least to make the initial array non-decreasing. If it is not possible with any amount of tricks, print $-1$ instead.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ($1\leq n \leq 10^6$, $1 \leq m \leq 10^6$) — the length of the initial array and the range of the elements in the array.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq m$) — the initial array.

The third line of each test case contains $m$ integers $b_1, \ldots, b_m$ ($1 \leq b_i \leq m$) — the fixed magic array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ and the sum of $m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the minimum number of tricks needed, or $-1$ if it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Examples:
input:
3
5 8
1 6 3 7 1
2 3 5 8 7 1 5 6
3 3
1 3 2
2 1 3
10 10
2 8 5 4 8 4 1 5 10 10
6 7 2 6 3 4 1 1 3 5
output:
3
-1
3

Note: In the first case, the initial array $a_1, \ldots, a_n$ is $[1, 6, 3, 7, 1]$. You can choose $S$ as follows:

  * first trick: $S = [2, 4, 5]$, $a = [1, 1, 3, 5, 2]$;    * second trick: $S = [5]$, $a = [1, 1, 3, 5, 3]$;    * third trick: $S = [5]$, $a = [1, 1, 3, 5, 5]$.

So it is possible to make $a_1, \ldots, a_n$ non-decreasing using $3$ tricks. It can be shown that this is the minimum possible amount of tricks.

In the second case, it is impossible to make $a_1, \ldots, a_n$ non- decreasing.

Buggy Code:
#include<bits/stdc++.h>
#define pb push_back
using namespace std;
const int N=1e6+5;
int T,n,m,x,a[N],b[N],f[N],dfn[N],d[N],sz[N],tot;
vector<int> ve[N];
inline int read(){
	int x=0;char c=getchar();
	while(c<'0'||c>'9')c=getchar();
	while(c>='0'&&c<='9')x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x;
}
int getf(int x){return x==f[x]?x:f[x]=getf(f[x]);}
void dfs(int u){dfn[u]=++tot;sz[u]=1;for(int v:ve[u])d[v]=d[u]+1,dfs(v),sz[u]+=sz[v];}
inline bool in(int x,int y){return dfn[x]>=dfn[y]&&dfn[x]<dfn[y]+sz[y];}
inline bool cgo(int x,int y,int k){return f[x]==f[y]&&((in(x,y)&&d[x]-d[y]<=k)||(f[x]==x&&in(b[x],y)&&d[b[x]]-d[y]<k));}
inline bool chk(int k){int x=1,y=1;while(x<=n&&y<=n)if(cgo(a[x],y,k))x++;else y++;return x>n;}
signed main(){
	T=read();
	while(T--){
		n=read(),m=read();tot=0;
		for(int i=1;i<=n;i++)a[i]=read();
		for(int i=1;i<=m;i++)f[i]=i;
		for(int i=1;i<=m;i++){x=read();if(getf(i)==getf(x))b[i]=x;else f[f[i]]=f[x],ve[x].pb(i);}
		for(int i=1;i<=m;i++)if(getf(i)==i)d[i]=1,dfs(i);
		int l=0,r=m;while(l<r){int mid=(l+r)>>1;if(chk(mid))r=mid;else l=mid+1;}
		if(!chk(m))puts("-1");else printf("%d\n",l);
		for(int i=1;i<=m;i++)ve[i].clear();
	}
	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
4 4
4 3 2 1
1 2 3 4
```