**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 800$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $800$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

Examples:
input:
6
5
1 5 3 2 4
8
2 1 3 4 5 6 7 8
3
1 2 3
4
1 2 4 4
5
4 4 1 3 5
1
1
output:
2
3
1
2
0
0

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
// LUOGU_RID: 164371228
#include<bits/stdc++.h>
using namespace std;
const int NR=805;
const int INF=0x3f3f3f3f;
int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
	return x*f;
}
int n;
int a[NR];
int dp[NR][NR],f[NR];
void work()
{
	n=read();
	for(int i=1;i<=n;i++)a[i]=read();
	memset(dp,0x3f,sizeof(dp));
	for(int i=1;i<=n+1;i++)dp[i][i-1]=0;
	for(int l=n;l>=1;l--)for(int r=l+1;r<=n;r++)
	{
		if(l<a[l]||(l-a[l]&1))continue;
		if(r==l+1){
			dp[l][r]=(l-a[l])/2;
			continue;
		}
		for(int i=l+1;i<=r;i++){
			int num=(l-a[l])/2;
			if(dp[l+1][i-1]<=num)dp[l][r]=max(num,max(dp[l+1][i-1],dp[i+1][r]-(i-l+1)/2));
		}
	}
	memset(f,0,sizeof(f));
	for(int i=0;i<=n;i++){
		f[i+1]=max(f[i],f[i+1]);
		for(int j=i+1;j<=n;j++){
			if(f[i]>=dp[i+1][j])f[j+1]=max(f[j+1],f[i]+(j-i)/2);
		}
	}
	printf("%d\n",f[n+1]);
}
int main()
{
	int T=read();while(T--)work();
	return 0;
}
/*
1
4
1 2 4 4
5
1 5 3 2 4
8
2 1 3 4 5 6 7 8
3
1 2 3
4
1 2 4 4
5
4 4 1 3 5
1
1
*/

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following points:

1. **Edge Cases**:
   - **Single Element Array**: The operation cannot be performed if the array has only one element.
   - **No Valid Operations**: The array may not have any elements that satisfy the condition \(a_i = i\).
   - **Multiple Valid Operations**: The array may have multiple elements that satisfy the condition \(a_i = i\), and the optimal sequence of operations needs to be carefully chosen.

2. **Complex Scenarios**:
   - **Overlapping Valid Operations**: The array may have multiple valid operations that overlap, making it tricky to determine the optimal sequence.
   - **Non-Contiguous Valid Operations**: The valid operations may be spread out, requiring careful consideration of the order in which they are performed.

3. **Buggy Code Analysis**:
   - The code uses dynamic programming (DP) to solve the problem, but it may have issues with how it handles the transitions and the conditions for valid operations.

Given these considerations, let's generate a test input that includes a mix of these scenarios:

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

This test input includes:
- A case with a single element (1).
- A case where no valid operations can be performed (4 4 1 3 5).
- A case with multiple valid operations that need careful sequencing (1 5 3 2 4).
- A case with a long sequence of valid operations (2 1 3 4 5 6 7 8).
- A simple case with a single valid operation (1 2 3).
- A case with overlapping valid operations (1 2 4 4).

This test input should expose the bug in the buggy code by challenging its ability to handle various edge cases and complex scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 800$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $800$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

Examples:
input:
6
5
1 5 3 2 4
8
2 1 3 4 5 6 7 8
3
1 2 3
4
1 2 4 4
5
4 4 1 3 5
1
1
output:
2
3
1
2
0
0

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
#include <bits/stdc++.h>
#define print(v) for (auto i:v) { cout << i << ' '; } cout << endl
#define ll long long
#define all(v) v.begin(),v.end()
using namespace std;

void solve() {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int i=0;i<n;i++) cin >> arr[i];
    int ans=0;
    while (arr.size()>1) {
        for (int j=arr.size()-2;j>=0;j--) {
            if (arr[j]==j+1) {
                arr.erase(arr.begin()+j,arr.begin()+j+2);
                ans++;
                break;
            }
            if (j==0) {
                cout << ans << '\n';
                return;
            }
        }
    }
    cout << ans << '\n';
}
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t=1;
    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
3
8
2 1 3 4 5 6 7 8
4
1 2 4 4
5
1 5 3 2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 800$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $800$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

Examples:
input:
6
5
1 5 3 2 4
8
2 1 3 4 5 6 7 8
3
1 2 3
4
1 2 4 4
5
4 4 1 3 5
1
1
output:
2
3
1
2
0
0

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
def func(s1,c,n,s2):
    s1 = s1+s2
    if n<=1:
        return c
    if n == 2 and s1[0]!=str(1):
        return c

    for i in range(n-2,-1,-1):
        if s1[i] == str(i+1):
            s2 = s1[i+2:]
            s1 = s1[:i]
            n=n-2
            c+=1
            break
        elif i == 0:
            return c
    # print(s1)
    # print(s2,c)
    return func(s1,c,n,s2)

test = int(input())
for t in range(test):
    n = int(input())
    s = list(map(str, input().split()))
    c=0

    ans = func(s,c,n,[])
    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:
To expose the bug in the given code, we need to focus on edge cases and scenarios that are likely to cause issues. The code provided has a few potential issues, such as:

1. **Handling of edge cases**: The code might not handle cases where the array is very short or has no valid operations.
2. **String vs. integer comparison**: The code converts the array elements to strings and compares them, which can lead to issues if the input is not handled correctly.
3. **Loop and recursion termination**: The loop and recursion might not terminate correctly in certain edge cases.

Let's consider a few test cases:

1. **Array with no valid operations**: An array where no elements match their indices.
2. **Array with a single valid operation**: An array where only one operation can be performed.
3. **Array with multiple valid operations**: An array where multiple operations can be performed, but the order of operations matters.

### Test Case 1: Array with no valid operations
```plaintext
1
5
4 5 1 3 2
```

### Test Case 2: Array with a single valid operation
```plaintext
1
4
1 3 2 4
```

### Test Case 3: Array with multiple valid operations
```plaintext
1
8
1 2 3 4 5 6 7 8
```

### Test Case 4: Edge case with very short array
```plaintext
1
1
1
```

### Test Case 5: Edge case with repeated elements
```plaintext
1
4
1 2 2 4
```

### Test Case 6: Edge case with all elements out of order
```plaintext
1
5
5 4 3 2 1
```

### Chosen Test Case
To maximize the failure probability, let's choose a test case where the array has multiple valid operations but the order of operations is crucial. This will test the code's ability to handle complex sequences correctly.

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

This test case will ensure that the code correctly identifies and performs multiple operations in the correct order. If the code fails to handle this case, it will likely expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 800$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $800$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

Examples:
input:
6
5
1 5 3 2 4
8
2 1 3 4 5 6 7 8
3
1 2 3
4
1 2 4 4
5
4 4 1 3 5
1
1
output:
2
3
1
2
0
0

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

Buggy Code:
import sys
input = sys.stdin.readline

for i in range(int(input())):
    n=int(input())
    l1=list(map(int,input().split()))
    ans=0
    while True:
        f1=False
        j=len(l1)-2
        while j>=0:
            if l1[j]==j+1:
                l1=l1[:j]+l1[j+2:]
                ans+=1
                f1=True
                break
            j-=1
        if f1==False:
            break
    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
3
5
1 2 3 4 5
6
1 2 3 4 5 6
7
1 2 3 4 5 6 7
```

### Explanation:
1. **Test Case 1:**
   - **Input:** `5 1 2 3 4 5`
   - **Output:** `2`
   - **Explanation:** The optimal sequence of operations is:
     - `[1, 2, 3, 4, 5]` → `[1, 3, 5]` (Remove 2 and 3)
     - `[1, 3, 5]` → `[5]` (Remove 1 and 3)
   - This test case checks if the code correctly handles a simple sequence where multiple operations can be performed.

2. **Test Case 2:**
   - **Input:** `6 1 2 3 4 5 6`
   - **Output:** `3`
   - **Explanation:** The optimal sequence of operations is:
     - `[1, 2, 3, 4, 5, 6]` → `[1, 3, 5, 6]` (Remove 2 and 3)
     - `[1, 3, 5, 6]` → `[1, 5]` (Remove 3 and 4)
     - `[1, 5]` → `[5]` (Remove 1 and 5)
   - This test case checks if the code can handle a slightly longer sequence with multiple operations.

3. **Test Case 3:**
   - **Input:** `7 1 2 3 4 5 6 7`
   - **Output:** `3`
   - **Explanation:** The optimal sequence of operations is:
     - `[1, 2, 3, 4, 5, 6, 7]` → `[1, 3, 4, 5, 7]` (Remove 2 and 3)
     - `[1, 3, 4, 5, 7]` → `[1, 4, 7]` (Remove 3 and 4)
     - `[1, 4, 7]` → `[7]` (Remove 1 and 4)
   - This test case checks if the code can handle a longer sequence with multiple operations and ensures that the code does not get stuck in an infinite loop.

These test cases are designed to expose any potential issues in the buggy code, such as incorrect handling of sequences where multiple operations can be performed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 800$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $800$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

Examples:
input:
6
5
1 5 3 2 4
8
2 1 3 4 5 6 7 8
3
1 2 3
4
1 2 4 4
5
4 4 1 3 5
1
1
output:
2
3
1
2
0
0

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

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

public class QN7 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while (T-- > 0) {
            int n = sc.nextInt();
            ArrayList<Integer> arr = new ArrayList<>(n);
            for (int i = 1; i <= n; i++) arr.add(sc.nextInt());

            int ans = 0;
            boolean flag = false;
            while (arr.size() > 1 && !flag) {
                flag = true;
                for (int i = arr.size() - 2; i >= 0; i--) {
                    if (arr.get(i) == i + 1) {
                        flag = false;
                        arr.remove(i + 1); // Remove the element at i + 1 first
                        arr.remove(i);     // Then remove the element at i
                        ans++;
                        break;
                    }
                }
            }
            System.out.println(ans);
        }
        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Interesting Problem (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference between the two versions is the constraint on $n$. You can make hacks only if both versions of the problem are solved.

You are given an array of integers $a$ of length $n$.

In one operation, you will perform the following two-step process:

  1. Choose an index $i$ such that $1 \le i < |a|$ and $a_i = i$.    2. Remove $a_i$ and $a_{i+1}$ from the array and concatenate the remaining parts.

Find the maximum number of times that you can perform the operation above.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 800$) — the length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $800$.

Output: For each test case, output a single integer — the maximum number of times that you can perform the operation.

Examples:
input:
6
5
1 5 3 2 4
8
2 1 3 4 5 6 7 8
3
1 2 3
4
1 2 4 4
5
4 4 1 3 5
1
1
output:
2
3
1
2
0
0

Note: In the first test case, one possible optimal sequence of operations is $[ 1, 5, \color{red}{3}, \color{red}{2}, 4 ] \rightarrow [\color{red}{1}, \color{red}{5}, 4] \rightarrow [4]$.

In the third test case, one possible optimal sequence of operations is $[1, \color{red}{2}, \color{red}{3}] \rightarrow [1]$.

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

public class App {
    public static void main(String[] args) throws Exception {
        Scanner input=new Scanner(System.in);
        int testCases=input.nextInt();
        int outputs[] = new int[testCases];
        ArrayList<ArrayList<Integer>> inputs = new ArrayList<>();
        for (int i=0; i<testCases; i++){
            int n=input.nextInt();
            ArrayList<Integer> testCase = new ArrayList<>();
            for(int j=0; j<n; j++)
            {
                testCase.add(input.nextInt());
            }
            inputs.add(testCase);
        }
        int outputInd = 0;
        for(ArrayList<Integer> a: inputs){
            int n = a.size();
            int ans = 0;
            for(int i=n-1; i>=1; i--){
                if(a.get(i-1) == i){
                    ans++;
                    a.remove(i);
                    a.remove(i-1);
                    n -= 2;
                    i = n;
                }
            }
            outputs[outputInd] = ans;
            outputInd++;
        }
        for(int output: outputs){
            System.out.println(output);
        }
        input.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:
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 can maximize the failure probability. The key issues in the buggy code are:

1. The code iterates from the end of the array to the beginning, which can cause issues if there are multiple valid indices to remove.
2. The code does not handle cases where the removal of elements might create new valid indices that were not previously valid.

Given these points, let's create a test input that can expose these issues:

### Test Input
```plaintext
3
5
1 2 3 4 5
6
1 2 3 4 5 6
7
1 2 3 4 5 6 7
```

### Explanation
1. **Test Case 1:**
   - Array: `[1, 2, 3, 4, 5]`
   - Expected Output: `2`
   - Explanation: The optimal sequence of operations is `[1, 2, 3, 4, 5] -> [1, 3, 5] -> [1]`.

2. **Test Case 2:**
   - Array: `[1, 2, 3, 4, 5, 6]`
   - Expected Output: `3`
   - Explanation: The optimal sequence of operations is `[1, 2, 3, 4, 5, 6] -> [1, 3, 5, 6] -> [1, 3] -> [1]`.

3. **Test Case 3:**
   - Array: `[1, 2, 3, 4, 5, 6, 7]`
   - Expected Output: `3`
   - Explanation: The optimal sequence of operations is `[1, 2, 3, 4, 5, 6, 7] -> [1, 3, 5, 7] -> [1, 3] -> [1]`.

### Why This Test Input is Challenging
1. **Test Case 1:**
   - The array has a sequence where multiple valid indices can be removed.
   - The code might fail to remove the correct indices in the optimal sequence.

2. **Test Case 2:**
   - The array has a longer sequence where multiple valid indices can be removed.
   - The code might fail to handle the removal of elements correctly, leading to incorrect results.

3. **Test Case 3:**
   - The array has an even longer sequence where multiple valid indices can be removed.
   - The code might fail to handle the removal of elements correctly, leading to incorrect results.

By providing these test cases, we can expose the issues in the buggy code and help identify the need for a more robust solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

Buggy Code:
#include<iostream>
using namespace std ;

int main()
{
    int t,n,k,x,y ;
    cin >> t ;

    while(t--){
        cin >> n >> k >> x >> y ; //x,y bodya & sasha starting position
        int p[n+1] ;
        int a[n+1] ;

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

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

        //bodya and sasha initial score
        unsigned long long int B = 0 ;
        unsigned long long int S = 0 ;

        int c = 1 ; //for a terminating purpose

        //total k turns,so loop run k times
        for(int i=0;i<k;i++){
            B+=a[x] ;

            while(c){
                if(a[p[x]] > a[x]){
                    x = p[x] ;
                    c = 1 ;
                }
                else
                    c = 0 ;
            }
        }

       // cout << "Bodya" << B << endl ;

        //for sasha

        c =1 ;

        for(int i=0;i<k;i++){
            S+=a[y] ;

            while(c){
                if(a[p[y]] > a[y]){
                    y = p[y] ;
                    c = 1 ;
                }
                else
                    c = 0 ;
            }
        }

        // cout << "Sasha" << S << endl ;

        if(B == S)
            cout << "Draw" << endl ;
        else if(B > S)
            cout << "Bodya" << endl ;
        else
            cout << "Sasha" << endl ;

    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

Buggy Code:
#include <iostream>
#include <vector>
using namespace std;

long long calculateScore(int pos, const pair<int, int>* arr, int n, long long k) {
    vector<bool> visited(n, false);
    vector<int> path;
    long long score = 0;
    int current_pos = pos - 1;
    long long moves = 0;

    while (moves < k) {
        if (visited[current_pos]) {
            int cycle_start = 0;
            for (int i = 0; i < path.size(); i++) {
                if (path[i] == current_pos) {
                    cycle_start = i;
                    break;
                }
            }

            int cycle_length = path.size() - cycle_start;
            long long cycle_score = 0;
            for (int i = cycle_start; i < path.size(); i++) {
                cycle_score += arr[path[i]].second;
            }

            long long remaining_moves = k - moves;
            long long complete_cycles = remaining_moves / cycle_length;
            long long leftover = remaining_moves % cycle_length;

            score += cycle_score * complete_cycles;
            for (int i = 0; i < leftover; i++) {
                score += arr[path[cycle_start + i]].second;
            }
            return score;
        }

        visited[current_pos] = true;
        path.push_back(current_pos);
        score += arr[current_pos].second;
        moves++;

        int next_pos = arr[current_pos].first - 1;
        if (next_pos < 0 || next_pos >= n ||
            arr[next_pos].second <= arr[current_pos].second) {
            score += arr[current_pos].second * (k - moves);
            break;
        }
        current_pos = next_pos;
    }

    return score;
}

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

    int t;
    cin >> t;

    while (t--) {
        int n;
        long long k;
        int bodya_start, sasha_start;
        cin >> n >> k >> bodya_start >> sasha_start;

        pair<int, int> arr[n];

        for (int i = 0; i < n; i++) {
            cin >> arr[i].first;
        }
        for (int i = 0; i < n; i++) {
            cin >> arr[i].second;
        }

        long long bodya_score = calculateScore(bodya_start, arr, n, k);
        long long sasha_score = calculateScore(sasha_start, arr, n, k);

        if (bodya_score > sasha_score) {
            cout << "Bodya\n";
        } else if (sasha_score > bodya_score) {
            cout << "Sasha\n";
        } else {
            cout << "Draw\n";
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
4 1000000000 1 2
2 3 4 1
1000000000 1000000000 1000000000 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

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

using namespace std;

long long getMaxScore(int *a, int *p, long long maxPoint, int startIndex, int k)
{

    long long maxScore = 0, currentIndex = startIndex - 1;
    long long currentScore = 0;
    for (int i = 0; i < k; i++)
    {
        currentScore += a[currentIndex];
        long long cScore = currentScore + (k - i - 1) * a[currentIndex];
        maxScore = max(maxScore, cScore);
        if (a[i] == maxPoint)
            break;
        currentIndex = p[currentIndex] - 1;
    }

    return maxScore;
}

void solve()
{
    int n, k, b, s;
    cin >> n >> k >> b >> s;
    int *p = new int[n];
    int *a = new int[n];

    for (int i = 0; i < n; i++)
        cin >> p[i];
    for (int i = 0; i < n; i++)
        cin >> a[i];

    int maxPoint = 0;
    for (int i = 0; i < n; i++)
        maxPoint = max(maxPoint, a[i]);

    long long bodyaScore = getMaxScore(a, p, maxPoint, b, k);
    long long sashaScore = getMaxScore(a, p, maxPoint, s, k);

    if (bodyaScore > sashaScore)
        cout << "Bodya" << endl;
    else if (sashaScore > bodyaScore)
        cout << "Sasha" << endl;
    else
        cout << "Draw" << endl;
}

int main()
{
    int t;
    cin >> t;
    // while(t-- > 0) solve();

    for (int i = 0; i < t; i++)
    {
        int n, k, b, s;
        cin >> n >> k >> b >> s;
        int *p = new int[n];
        int *a = new int[n];

        for (int i = 0; i < n; i++)
            cin >> p[i];
        for (int i = 0; i < n; i++)
            cin >> a[i];

          if(t== 10000 && i != 183) return 0;
        if (t == 10000 && i == 183)
        {
            cout << n << " " << k << " " << b << " " << s << endl;
            for (int i = 0; i < n; i++)
                cout << p[i] << " ";
            cout << endl;
            for (int i = 0; i < n; i++)
                cout << a[i] << " ";
            return 0;
        }

        int maxPoint = 0;
        for (int i = 0; i < n; i++)
            maxPoint = max(maxPoint, a[i]);

        long long bodyaScore = getMaxScore(a, p, maxPoint, b, k);
        long long sashaScore = getMaxScore(a, p, maxPoint, s, k);

        if (bodyaScore > sashaScore)
            cout << "Bodya" << endl;
        else if (sashaScore > bodyaScore)
            cout << "Sasha" << endl;
        else
            cout << "Draw" << endl;
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

Buggy Code:
for _ in range(int(input())):
    n, k, pb, ps = map(int, input().split())
    p = [0]+list(map(int, input().split()))
    a = [0]+list(map(int, input().split()))
    sb = []
    ss = []
    ipb = pb
    ips = ps
    for _ in range(k):
        sb.append(a[pb])
        pb = p[pb]
        if pb == ipb: break
    for _ in range(k):
        ss.append(a[ps])
        ps = p[ps]
        if ps == ips: break
    lb = len(sb)
    ls = len(ss)
    mb = max(sb)
    ms = max(ss)
    dsb = (sum(sb)*(k//lb-1))+sum(sb[:k%lb])
    dss = (sum(ss)*(k//ls-1))+sum(ss[:k%ls])
    dbi1 = [sum(sb[i:])+dsb for i in range(lb)]
    dsi1 = [sum(ss[i:])+dss for i in range(ls)]
    dbi2 = [sb[i]*(k-i) for i in range(lb)]
    dsi2 = [ss[i]*(k-i) for i in range(ls)]
    dbi = [max(dbi1[i], dbi2[i]) for i in range(lb)]
    dsi = [max(dsi1[i], dsi2[i]) for i in range(ls)]

    # continue
    kb = k
    i = 0
    while i < kb:
        if sb[i] == mb:
            kb = i+1
            break
        if sb[i]*(kb-i) > dbi[i+1]:
            kb = i+1
            break
        if sb[i%lb] >= sb[(kb-1)%lb]: kb -= 1
        else: i += 1
    ks = k
    i = 0
    while i < ks:
        if ss[i%ls] == ms:
            ks = i+1
            break
        if ss[i]*(ks-i) > dsi[i+1]:
            ks = i+1
            break
        if ss[i%ls] >= ss[(ks-1)%ls]: ks -= 1
        else: i += 1
    # print(sb[:kb], ss[:ks])
    sb = sum(sb[i%lb] for i in range(kb)) + sb[kb-1] * (k-kb)
    ss = sum(ss[i%ls] for i in range(ks)) + ss[ks-1] * (k-ks)
    # print(sb, ss)
    if sb > ss: print("Bodya")
    elif ss > sb: print("Sasha")
    else: print("Draw")

**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 1 3
2 3 4 5 1
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: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

Buggy Code:
def solve():
    n, k, pb, ps = map(int, input().split())
    pb -= 1  # Convert indexing to 0-based
    ps -= 1  # Convert indexing to 0-based

    p = [int(x) - 1 for x in input().split()]  # Read and convert player indices
    a = [int(x) for x in input().split()]

    max_strength = max(a);
    maxB = 0
    smB=0
    maxS = 0
    smS = 0
    for i in range(min(k,n)):
        smB+= a[pb]
        smS += a[ps]
        maxS = max(maxS, smS + a[ps]*(k-i-1))
        maxB = max(maxB, smB + a[pb]*(k-i-1))
        if(a[pb] == max_strength or a[ps] == max_strength): break
        pb = p[pb]
        ps = p[ps]

    if(maxB>maxS): print("Bodya")
    elif(maxS>maxB): print("Sasha")
    else: print("Draw")

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

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

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

Think step by step.

Response:
```plaintext
1
4 10 1 3
2 3 4 1
10 1 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

Buggy Code:
import sys

first = True

count = 0

def permutation_game(n, k, p, a, PB, PS):
    # Adjust indices for 0-based array manipulation
    PB -= 1
    PS -= 1

    def simulate_game(start, turns):
        current = start
        score = 0
        visited = {}
        history_scores = []

        for t in range(turns):
            if current in visited:
                # Loop detected, compute scores in the loop efficiently
                loop_start = visited[current]
                loop_length = t - loop_start
                loop_score = sum(history_scores[loop_start:t])
                remaining_turns = turns - t

                # Full loops we can execute
                full_loops = remaining_turns // loop_length
                score += full_loops * loop_score

                # Remaining part of the loop
                remainder = remaining_turns % loop_length
                score += sum(history_scores[loop_start:loop_start+remainder])
                break

            # Record this position and the turn it was visited
            visited[current] = t
            history_scores.append(a[current])
            score += a[current]

            # Check both staying and moving to next position
            stay_score = a[current]
            move_score = a[p[current] - 1]  # convert to 0-based index

            if stay_score > move_score:
                # Staying is better or the same as moving
                next_pos = current
            else:
                # Moving is better
                next_pos = p[current] - 1

            if next_pos != current:
                current = next_pos
            else:
                # if the next position is the current one, stay and keep collecting score
                score += a[current] * (turns - t - 1)
                break

        return score

    # Get scores for both players
    score_bodya = simulate_game(PB, k)
    score_sasha = simulate_game(PS, k)

    # Determine the result
    if score_bodya > score_sasha:
        return "Bodya"
    elif score_sasha > score_bodya:
        return "Sasha"
    else:
        return "Draw"

for line in sys.stdin:
    if first:
        first = False
        n = int(line.strip())
        continue

    if count % 3 == 0:
        n, k, pb, ps = map(int, line.strip().split())
    elif count % 3 == 1:
        p = list(map(int, line.strip().split()))
    else:
        a = list(map(int, line.strip().split()))
        k = permutation_game(n, k, p, a, pb, ps)
        print(k)
    count += 1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class b1{
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int testCase = Integer.parseInt(br.readLine());
        StringBuilder sb = new StringBuilder();
        while(testCase-->0){
            StringTokenizer st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int k = Integer.parseInt(st.nextToken());
            int pb = Integer.parseInt(st.nextToken());
            int ps = Integer.parseInt(st.nextToken());
            int[] pArray = new int[n+1];
            boolean[] visitedB = new boolean[n+1];
            boolean[] visitedS = new boolean[n+1];
            st = new StringTokenizer(br.readLine());
            for(int i=1; i<=n; i++){
                pArray[i] = Integer.parseInt(st.nextToken());
            }
            st = new StringTokenizer(br.readLine());
            int[] scores = new int[n+1];
            for(int i=1; i<=n; i++){
                scores[i] = Integer.parseInt(st.nextToken());
            }

            long maxBScore =  (long) scores[pb] * k;
            long currentBScore = 0;

            long maxSScore = (long) scores[ps] * k;
            long currentSScore = 0;

            for(int i=0; i<k; i++){
                currentSScore += scores[ps];
                visitedS[ps] = true;
                ps = pArray[ps];
                maxSScore = Math.max(maxSScore, currentSScore + ((long)scores[ps] * (k-i-1)));
                if(visitedS[ps]) break;

                currentBScore += scores[pb];
                visitedB[pb] = true;
                pb = pArray[pb];
                maxBScore = Math.max(maxBScore, currentBScore + ((long)scores[pb] * (k-i-1)));
                if(visitedB[pb]) break;
            }

            if(maxBScore == maxSScore) {
                sb.append("Draw");
            }
            else if(maxBScore > maxSScore) {
                sb.append("Bodya");
            }
            else {
                sb.append("Sasha");
            }
            sb.append('\n');

        }
        System.out.println(sb);

    }
}

**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 3 1 2
2 3 4 1
5 1 6 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

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

public class CP {

    static final int M = (int) 1e9 + 7;
    static final int INT_INF = 2_000_000_000;
    static final long LONG_INF = 9_000_000_000_000_000_000L;
    static final int N = (int) 2e6;
    static long[] factorial;
    static final boolean[] isPrime = new boolean[N + 1];
    static final ArrayList<Integer> primes = new ArrayList<>();
    static final Random random = new Random();
    static final FastScanner fs = new FastScanner();
    static final FastOutput fo = new FastOutput();

    //variable

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

        int t = fs.nextInt();

        while (t-- > 0) {

            int n = fs.nextInt();
            long k = fs.nextInt();

            int pb = fs.nextInt() - 1;
            int ps = fs.nextInt() - 1;
            int[] p = fs.arrayIn(n);
            int[] a = fs.arrayIn(n);

            long dpB = a[pb];
            long dpS = a[ps];

            for(int i=1; i<k; i++) {
                if(dpB + a[pb] < dpB + a[p[pb] - 1]) {
                    dpB = dpB + a[p[pb] - 1];
                    pb = p[pb] - 1;
                } else {
                    dpB = dpB + a[pb];
                    dpB += (k - i) * a[pb];
                    break;
                }
            }

            for(int i=1; i<k; i++) {
                if(dpS + a[ps] < dpS + a[p[ps] - 1]) {
                    dpS = dpS + a[p[ps] -1];
                    ps = p[ps] -1;
                } else {
                    dpS = dpS + a[ps];
                    dpS += (k - i) * a[ps];
                    break;
                }
            }

            if(dpS < dpB) {
                fo.println("Bodya");
            } else if (dpB < dpS) {
                fo.println("Sasha");
            } else {
                fo.println("Draw");
            }
        }
        fo.flush();
    }

    //class

    //function
    private static boolean check(char[][] grid, int n, int m, int s, int e) {
        char c = grid[s][e];
        byte check = 0;
        for (int i = 0; i < n; i++) {
            if (c != grid[i][e]) {
                check++;
                break;
            }
        }

        for (int i = 0; i < m; i++) {
            if (c != grid[s][i]) {
                check++;
                break;
            }
        }

        return check == 2;
    }

    private static boolean isPalindrome(String s) {
        int n = s.length(), i = 0;
        while (i < n / 2) {
            if (s.charAt(i) != s.charAt(n - i - 1)) return false;
            i++;
        }

        return true;
    }

    private static String test(int t, int n) {
        StringBuilder testCase = new StringBuilder();
        //testCase.append(t);
        // testCase.append("\n");
        while (t-- > 0) {
            int a = random.nextInt(n);
            testCase.append(a);
            testCase.append(" ");
            a = random.nextInt(a);
            testCase.append(a);
            testCase.append("\n");
        }

        return testCase.toString();
    }

    //Template
    static void getPrimes() {
        for (int i = 2; i < N; i++) isPrime[i] = true;

        int i = 2;
        while (i < N) {
            if (isPrime[i]) {
                primes.add(i);
                for (int j = 2 * i; j < N; j += i)
                    isPrime[j] = false;
            }
            i++;
        }
    }

    static void build(int[] a, int[] seg, int ind, int low, int high) {
        if (low == high) {
            seg[ind] = a[low];
            return;
        }

        int mid = (low + high) / 2;
        build(a, seg, 2 * ind + 1, low, mid);
        build(a, seg, 2 * ind + 2, mid + 1, high);

        seg[ind] = Math.max(seg[2 * ind + 1], seg[2 * ind + 2]);
    }

    static long query(int ind, int[] seg, int l, int h, int low, int high) {
        if (low > h || high < l) return -INT_INF;

        if (low >= l && high <= h) return seg[ind];

        int mid = (low + high) / 2;
        long left = query(2 * ind + 1, seg, l, h, low, mid);
        long right = query(2 * ind + 2, seg, l, h, mid + 1, high);

        return Math.max(left, right);
    }

    static long factorial(int n) {
        long fact = 1;
        for (int i = 1; i <= n; i++) {
            fact *= i;
        }
        return fact;
    }

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

    static long gcd(long a, long b) {
        if (a == 0) return b;

        return gcd(b % a, a);
    }

    static void permutation(int n, ArrayList<Integer> arr, boolean[] chosen) {
        if (arr.size() == n) {
//            System.out.println(arr);
        } else {
            for (int i = 1; i <= n; i++) {
                if (chosen[i]) continue;
                arr.add(i);
                chosen[i] = true;
                permutation(n, arr, chosen);
                arr.remove(Integer.valueOf(i));
                chosen[i] = false;
            }
        }
    }

    static boolean isPalindrome(char[] c) {
        int n = c.length;
        for (int i = 0; i < n / 2; i++) {
            if (c[i] != c[n - i - 1]) return false;
        }
        return true;
    }

    static long nCk(int n, int k) {
        return (modMult(fact(n), fastExpo(modMult(fact(n - k), fact(k)), M - 2)));
    }

    static long nPk(int n, int k) {
        return (modMult(fact(n), fastExpo(fact(n - k), M - 2)));
    }

    static long fact(int n) {

        if (factorial != null) return factorial[n];
        else factorial = new long[N + 1];

        factorial[0] = 1;
        long fact = 1;
        for (int i = 1; i <= N; i++) {
            factorial[i] = fact = modMult(fact, i);
        }

        return factorial[n];
    }

    static long modMult(long a, long b) {
        return (int) (a * b % M);
    }

    static long negMult(long a, long b) {
        return (int) ((a * b) % M + M) % M;
    }

    static long fastExpo(long x, int y) {
        if (y == 1) return x;
        if (y == 0) return 1;
        long ans = fastExpo(x, y / 2);
        if (y % 2 == 0) return modMult(ans, ans);
        else return modMult(ans, modMult(ans, x));
    }

    static void ruffleSort(int[] arr) {
        int n = arr.length;

        for (int i = 0; i < n; i++) {
            int j = random.nextInt(n);
            int temp = arr[j];
            arr[j] = arr[i];
            arr[i] = temp;
        }

        Arrays.sort(arr);
    }

    static void ruffleSort(long[] arr) {
        int n = arr.length;

        for (int i = 0; i < n; i++) {
            int j = random.nextInt(n);
            long temp = arr[j];
            arr[j] = arr[i];
            arr[i] = temp;
        }

        Arrays.sort(arr);
    }

    private static class Pair implements Comparable<Pair> {
        int f, s;

        Pair(int f, int s) {
            this.f = f;
            this.s = s;
        }

        public int compareTo(Pair p) {
            if (this.f != p.f) return Integer.compare(this.f, p.f);

            return Integer.compare(this.s, p.s);
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (!(o instanceof Pair)) return false;
            Pair pairs = (Pair) o;
            return f == pairs.f && s == pairs.s;
        }

        @Override
        public int hashCode() {
            return Objects.hash(f, s);
        }

        @Override
        public String toString() {
            return "Pair: " + this.f + ", " + this.s;
        }
    }

    private static class FastScanner {
        //                Reader inputString = new StringReader(test(1, 1000));
//        BufferedReader br = new BufferedReader(inputString);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer str = new StringTokenizer("");

        String next() throws IOException {
            while (!str.hasMoreTokens()) {
//                String line = br.readLine();
//                str = new StringTokenizer(line);
//                System.out.println(line);
                str = new StringTokenizer(br.readLine());
            }

            return str.nextToken();
        }

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

        char nextChar() throws IOException {
            return next().charAt(0);
        }

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

        float nextFloat() throws IOException {
            return Float.parseFloat(next());
        }

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

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

        byte nextByte() throws IOException {
            return Byte.parseByte(next());
        }

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

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

    private static class FastOutput {
        private static final BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));

        StringBuilder output = new StringBuilder();

        void println(String... s) throws IOException {
            for (String a : s) output.append(a).append(" ");
            output.append("\n");
        }

        void print(int i) throws IOException {
            output.append(i);
        }

        void println(int... i) throws IOException {
            for (int a : i) output.append(a).append(" ");
            output.append("\n");
        }

        void println(char c) throws IOException {
            output.append(c).append("\n");
        }

        void println(long... l) throws IOException {
            for (long a : l) output.append(a).append(" ");
            output.append("\n");
        }

        void println(double... d) throws IOException {
            for (double a : d) output.append(a).append(" ");
            output.append("\n");
        }

        void println(float... f) throws IOException {
            for (float a : f) output.append(a).append(" ");
            output.append("\n");
        }

        void flush() throws IOException {
            out.write(output.toString());
            out.flush();
        }
    }
}

**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 that might cause the algorithm to fail. The primary issue in the provided code is the way it handles the movement and score calculation. Specifically, the code incorrectly decrements the indices when moving to the next position in the permutation, which can lead to out-of-bounds errors or incorrect score calculations.

Let's create a test case that highlights this issue:

1. **Small Permutation and Array Size**: Use a small permutation and array to make the test case manageable and easy to analyze.
2. **Starting Positions**: Choose starting positions that will lead to the bug being triggered.
3. **Permutation and Array Values**: Set values that will cause the bug to be evident in the score calculation.

Here's a test input that should expose the bug:

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

### Explanation:
- **Permutation**: `4 1 2 3`
  - This permutation means:
    - `p[0] = 4` (which is out of bounds if not properly handled)
    - `p[1] = 1`
    - `p[2] = 2`
    - `p[3] = 3`
- **Array `a`**: `7 2 5 6`
  - This array will be used to calculate scores.
- **Starting Positions**: `P_B = 3` (Bodya starts at position 3), `P_S = 2` (Sasha starts at position 2)
- **Number of Turns**: `k = 2`

### Expected Behavior:
- **Bodya's Moves**:
  - Start at position 3, score = `a[3] = 6`
  - Move to `p[3] = 3`, score = `6 + 6 = 12`
- **Sasha's Moves**:
  - Start at position 2, score = `a[2] = 5`
  - Move to `p[2] = 2`, score = `5 + 5 = 10`

### Actual Behavior:
- The code incorrectly decrements the indices when moving to the next position, which can lead to out-of-bounds errors or incorrect score calculations. For example, `p[3] - 1` will be 2, which is incorrect.

This test case should help expose the bug in the code by causing it to fail due to incorrect index handling.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
typedef long long LL;
int a[N],p[N];
void solve()
{
    int n,k,pb,ps;
    cin>>n>>k>>pb>>ps;
    map<int,int> mp;
    for(int i=1;i<=n;i++) cin>>p[i];
    for(int i=1;i<=n;i++) cin>>a[i];
    priority_queue<LL> Bo,Sa;
    LL sumb=0,sums=0;
    int pb1=pb,ps1=ps;
    int k1=min(k,2*n);
        for(int i=1;i<=k1;i++)
        {
            if(a[pb1]<a[pb]) Bo.push(a[pb]*(k1-i+1)+sumb);
            if(a[ps1]<a[ps]) Sa.push(a[ps]*(k1-i+1)+sums);
            sumb+=a[pb1];
            sums+=a[ps1];
            Bo.push(sumb);
            Sa.push(sums);
            pb=pb1,ps=ps1;
            pb1=p[pb1],ps1=p[ps1];
        }
    LL tb=Bo.top(),ta=Sa.top();
    if(tb>ta) cout<<"Bodya\n";
    else if(tb<ta) cout<<"Sasha\n";
    else cout<<"Draw\n";
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    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
4 2 3 2
4 1 2 3
7 2 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bodya and Sasha found a permutation $$$p_1,\dots,p_n$$$ and an array $$$a_1,\dots,a_n$$$. They decided to play a well-known "Permutation game".

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Both of them chose a starting position in the permutation.

The game lasts $$$k$$$ turns. The players make moves simultaneously. On each turn, two things happen to each player:

- If the current position of the player is $$$x$$$, his score increases by $$$a_x$$$.
- Then the player either stays at his current position $$$x$$$ or moves from $$$x$$$ to $$$p_x$$$.

Knowing Bodya's starting position $$$P_B$$$ and Sasha's starting position $$$P_S$$$, determine who wins the game if both players are trying to win.

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

The first line of each testcase contains integers $$$n$$$, $$$k$$$, $$$P_B$$$, $$$P_S$$$ ($$$1\le P_B,P_S\le n\le 2\cdot 10^5$$$, $$$1\le k\le 10^9$$$) — length of the permutation, duration of the game, starting positions respectively.

The next line contains $$$n$$$ integers $$$p_1,\dots,p_n$$$ ($$$1 \le p_i \le n$$$) — elements of the permutation $$$p$$$.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$1\le a_i\le 10^9$$$) — elements of array $$$a$$$.

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

Output: For each testcase output:

- "Bodya" if Bodya wins the game.
- "Sasha" if Sasha wins the game.
- "Draw" if the players have the same score.

Examples:
input:
10
4 2 3 2
4 1 2 3
7 2 5 6
10 8 2 10
3 1 4 5 2 7 8 10 6 9
5 10 5 1 3 7 10 15 4 3
2 1000000000 1 2
1 2
4 4
8 10 4 1
5 1 4 3 2 8 6 7
1 1 2 1 2 100 101 102
5 1 2 5
1 2 4 5 3
4 6 9 4 2
4 2 3 1
4 1 3 2
6 8 5 3
6 9 5 4
6 1 3 5 2 4
6 9 8 9 5 10
4 8 4 2
2 3 4 1
5 2 8 7
4 2 3 1
4 1 3 2
6 8 5 3
2 1000000000 1 2
1 2
1000000000 2
output:
Bodya
Sasha
Draw
Draw
Bodya
Sasha
Sasha
Sasha
Sasha
Bodya

Note: Below you can find the explanation for the first testcase, where the game consists of $$$k=2$$$ turns.

TurnBodya's positionBodya's scoreBodya's moveSasha's positionSasha's scoreSasha's movefirst$$$3$$$$$$0 + a_3 = 0 + 5 = 5$$$stays on the same position$$$2$$$$$$0 + a_2 = 0 + 2 = 2$$$moves to $$$p_2=1$$$second$$$3$$$$$$5 + a_3 = 5 + 5 = 10$$$stays on the same position$$$1$$$$$$2 + a_1 = 2 + 7 = 9$$$stays on the same positionfinal results$$$3$$$$$$10$$$$$$1$$$$$$9$$$

As we may see, Bodya's score is greater, so he wins the game. It can be shown that Bodya always can win this game.

Buggy Code:
#include <bits/stdc++.h>
#include <iostream>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set1;
typedef tree<int, null_type, greater<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set2;
typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update> ordered_set_s;
typedef tree<pair<int, int>, null_type, less_equal<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update> ordered_set_s2;
typedef long long int ll;
#define fast                      \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
const int N = 1200300;
#define pi 2 * acos(0.0)
#define vll vector<ll>
#define pq_ priority_queue
#define pq_min <int,vector<int>,greater<int>>
#define vi vector<int>
#define vii vector<pair<int, int>>
#define pii pair<int, int>
#define pb push_back
#define sp ' '
#define sz(n) (int)n.size()
#define all(n) n.begin(), n.end()
#define rall(n) n.rbegin(), n.rend()
#define YES cout << "YES\n";
#define yes cout << "yes\n";
#define NO cout << "NO\n";
#define no cout << "no\n";
#define bye return 0;
#define ss second;
#define ff first;
const double eps = 1e-9;
const int MAX = 1e5;
const int lim = 1e6;
bool odd(ll num) { return ((num & 1) == 1); }
bool even(ll num) { return ((num & 1) == 0); }
bool isEqual(double a, double b) { return abs(a - b) < eps; }
bool isGreater(double a, double b) { return a >= b + eps; }
bool isGreaterEqual(double a, double b) { return a > b - eps; }
#define tc     \
    int tt;    \
    cin >> tt; \
    while (tt--)
#define mset(n, v) memset(n, v, sizeof(n))
#define chk(n) for (auto it : n)
#define ff1(i, n) for (int i = 1; i <= n; i++)
#define lcm(a, b) (((a) * (b)) / __gcd(a, b))
int dx[8] = {1, -1, 0, 0, 1, -1, -1, 1};
int dy[8] = {0, 0, 1, -1, 1, -1, 1, -1};
ll mn(ll x, ll y)
{
    if (x < y)
        return x;
    else
        return y;
}
ll maxx(ll x, ll y)
{
    if (x > y)
        return x;
    else
        return y;
}std::vector<int> bin(int n) {
    std::vector<int> binary;

    while (n > 0) {
        binary.push_back(n % 2);
        n /= 2;
    }

    // Reverse the vector to get the binary representation in correct order
    std::reverse(binary.begin(), binary.end());

    return binary;
}

int ispal(int n)
{

    vi v = bin(n);
    int f = 0;
    for (int i = 0, j = v.size() - 1; i < v.size() / 2; i++, j--)
    {
        if (v[i] != v[j])
            f = 1;
    }
    if (!f)
        return 1;
    return 0;
}

ll POW(ll a, ll b)
{
    if (!b)
        return 1;
    ll r = POW(a, b / 2);
    if (b % 2)
        return r * r * a;
    else
        return r * r;
}
void pri(vi v)
{
    for (auto it : v)
        cout << it << ' ';
    cout << endl;
}
void pro(vll v)
{
    for (auto it : v)
        cout << it << ' ';
    cout << endl;
}
int mex(vi v, int l, int r)
{
    vi a;
    for (int i = l; i <= r; i++)
    {
        a.pb(v[i]);
    }
    sort(all(a));
    for (int i = 0, j = 1; i < a.size(); i++)
    {
        if (a[i] == j)
        {
            j++;
        }
        else
        {
            return j;
        }
    }
    return a.back() + 1;
}
vector<int> cps(const vector<int> &arr)
{
    int n = arr.size();
    vector<int> prefixSum(n);

    if (n > 0)
    {
        prefixSum[0] = arr[0];
        for (int i = 1; i < n; ++i)
        {
            prefixSum[i] = prefixSum[i - 1] + arr[i];
        }
    }

    return prefixSum;
}

bool comp(pair<int, int> x, pair<int, int> y)
{
    if (x.second < y.second)
        return true;
    else if (x.second == y.second)
    {
        if (x.first > y.first)
            return true;
        else
            return false;
    }
    else
        return false;
}
ll PoW(ll base, ll exponent) {
    ll result = 1;
    for (int i = 0; i < exponent; ++i) {
        result *= base;
        if (result > 1000000) {
            return -1;
        }
    }
    return result;
}

bool isp(ll num) {
    ll om = num;
    ll rm = 0;
    while(num>0) {
        ll digit = num % 10;
        rm = rm * 10 + digit;
        num /= 10;
    }
    return om == rm;
}
string lcs(const std::vector<std::string>& strs) {
    std::vector<std::string::const_reverse_iterator> backs;
    std::string s;
    if (strs.size() == 0) return "";
    if (strs.size() == 1) return strs[0];
    for (auto& str : strs) backs.push_back(str.crbegin());
    while (backs[0] != strs[0].crend()) {
        char ch = *backs[0]++;
        for (std::size_t i = 1; i<strs.size(); i++) {
            if (backs[i] == strs[i].crend()) goto done;
            if (*backs[i] != ch) goto done;
            backs[i]++;
        }
        s.push_back(ch);
    }
done:
    reverse(s.begin(), s.end());
    return s;
}
bool leap(int year) {
    return (year % 4 == 0 && (year % 100 != 0 || year % 400 == 0));
}
bool isd(string s,ll d , ll y)
{
    if(d==32) return 0;
    if(s=="April"|| s=="June"|| s=="September"|| s=="November")
    {
        if(d==31) return 0;
        return 1;
    }
    if(d>28 && s=="February")
    {
         if(d==29 && leap(y))
         {
            return 1;
         }
         return 0;
    }
    return 1;
}
vll pref(vll v)
{
    vll pr;
    pr.pb(v.front());
    for(int i=1;i<v.size();i++)
    {
        pr.pb(v[i]+pr[i-1]);
    }
    return pr;
}
ll cxor(vi v, ll l,ll r)
{
    ll xo=v[l];
    for(int i=l+1;i<=r;i++)
    {
        xo^=v[i];
    }
    return xo;
}
#define seea(a,x,y) for(int i=x;i<y;i++){cin>>a[i];}
#define seev(v,n) for(int i=0;i<n;i++){int x; cin>>x; v.push_back(x);}
#define sees(s,n) for(int i=0;i<n;i++){int x; cin>>x; s.insert(x);}
int cn(const vector<int>& arr) {
    int count = 0;
    int n = arr.size();
    ll prev=-1;
    for (int i = 0; i < n; ++i) {
        if (arr[i] % 2 == 0) {
            count += i + 1;
            if(prev!=-1)
            count += i - prev ;

            prev = i;
        }
    }
    cout<<endl;

    return count;
}

int main(){
    tc{
        ll n,k,pd,ps;
        cin>>n>>k>>pd>>ps;
        ll nn=n;
        vll p,a;
        seev(p,n);
        seev(a,n);
        ll mx=*max_element(all(a));
        pd--;
        ps--;
        ll s1=0,s2=0;
        vll p1,p2;;

        while(nn--){

            p1.pb(pd);
            p2.pb(ps);

            pd=p[pd]-1;
            ps=p[ps]-1;
        }
        vll val1,val2;
        for(int i=0;i<min(n,k);i++)// path of a
        {
            val1.pb(a[p1[i]]);
            val2.pb(a[p2[i]]);
        }
        ll x=val1.size(),sum1=accumulate(all(val1),0ll);
        ll sum2=accumulate(all(val2),0ll);
        map<ll,ll>m1,m2;
        ll o=0,q=0;
        for(int i=0;i<x;i++){
            o+=val1[i];
            q+=val2[i];
            m1[i]=o+(x-i-1)*val1[i] ;
            m2[i]=q+(x-i-1)*val2[i] ;
        }

        ll pos1=-1, pos2=-1;
        s1=sum1,s2=sum2;// just taken all
        for(auto it: m1){

                if(it.second>=s1){
                    s1=it.second;// staying at this position
                    pos1=it.first;
                }

        }
         for(auto it: m2){

                if(it.second>=s2){
                    s2=it.second;// staying at this position
                    pos2=it.first;
                }

        }
        ll pst=0;
        for(int i=0;i<n;i++){
            if(a[i]==mx) pst=i;;
        }

        if(k>n){
            k-=n;
            if(pos1!=-1)
            s1+=val1[pos1]*k;
            else
            {
                ll d=k/n,mod=k%n;
                s1+=d*sum1;
                for(int i=0;i<mod;i++){
                    s1+=val1[i];
                }
            }
            if(pos2!=-1)
            s2+=val2[pos2]*k;
            else
            {
                ll d=k/n,mod=k%n;
                s2+=d*sum2;
                for(int i=0;i<mod;i++){
                    s2+=val2[i];
                }
            }
        }
        // cout<<s1<<' '<<s2<<endl;
        string B="Bodya", S="Sasha", D="Draw";
        if(s1>s2) cout<<B<<endl;
        else if(s2>s1) cout<<S<<endl;
        else cout<<D<<endl;

    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's consider the following aspects:

1. **Edge Cases:**
   - Small values of `n` and `k` to ensure the test case is small.
   - Values of `P_B` and `P_S` that are the same or very close to each other.
   - Values of `P_B` and `P_S` that lead to different cycles in the permutation.

2. **Challenging Scenarios:**
   - Ensure that the permutation and array `a` have values that can lead to different scores when players stay or move.
   - Consider cases where the maximum value in `a` is in a position that can be reached by both players but with different paths.

Let's construct a test case with these considerations:

- **n = 4**: Small permutation size.
- **k = 2**: Short game duration.
- **P_B = 1**: Starting position for Bodya.
- **P_S = 2**: Starting position for Sasha.
- **p = [2, 3, 4, 1]**: Permutation that forms a cycle.
- **a = [10, 20, 30, 40]**: Array with increasing values to differentiate scores.

Here is the test input:

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

### Explanation:
- **Turn 1:**
  - Bodya starts at position 1, score = 10.
  - Sasha starts at position 2, score = 20.
  - Bodya moves to position 2, score = 10 + 20 = 30.
  - Sasha moves to position 3, score = 20 + 30 = 50.

- **Turn 2:**
  - Bodya is at position 2, score = 30 + 20 = 50.
  - Sasha is at position 3, score = 50 + 30 = 80.
  - Bodya moves to position 3, score = 50 + 30 = 80.
  - Sasha moves to position 4, score = 80 + 40 = 120.

- **Final Scores:**
  - Bodya: 80
  - Sasha: 120

In this case, Sasha wins the game. This test case should expose the bug in the buggy code by showing that the scores are calculated correctly and that Sasha wins.