**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=100010;
int main()
{
	int t;cin>>t;
	while(t--)
	{
		int n;cin>>n;
		int a[N];
		int z=1000001;
		int biao=1;
		for(int i=1;i<=n;i++)
		{
			cin>>a[i];
			if(a[i]<z)
			{
				z=a[i];
				biao=i;
			}
		}
		if(a[1]==a[n])
		{
			int m=n;
			while(a[n]==a[m])m--;
			biao=m+1;
		}
		int x=0,y=0;
		int flag=1;
		for(int i=biao;i<=biao+n-2;i++)
		{
			if(i>=n)
			{
				x=i-n+1;
				if(i>n)y=i-n;
				else y=i;
			}
			else
			{
				x=i+1;
				y=i;
			}
			if(a[y]>a[x])
			{
				flag=0;
				break;
			}
		}
		if(flag==0)cout<<"NO"<<endl;
		else cout<<"YES"<<endl;
	}
	return 0;
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1:**
   - Array: `[1, 3, 2, 4, 5]`
   - This array cannot be sorted into non-decreasing order by any number of prefix-suffix swaps. The elements `3` and `2` are out of order and cannot be fixed by a single swap.

2. **Test Case 2:**
   - Array: `[1, 2, 3, 4, 5, 6]`
   - This array is already sorted, so it can be considered as already non-decreasing. No swaps are needed.

3. **Test Case 3:**
   - Array: `[4, 3, 2, 1]`
   - This array can be sorted by performing a single prefix-suffix swap. If we split the array into `x = [4, 3, 2]` and `y = [1]`, swapping them results in `y + x = [1, 4, 3, 2]`. Then, if we split it again into `x = [1]` and `y = [4, 3, 2]`, swapping them results in `y + x = [4, 3, 2, 1]` which is already sorted.

These test cases cover different scenarios:
- An array that cannot be sorted.
- An already sorted array.
- An array that can be sorted with a single swap.

These inputs are designed to challenge the buggy code and expose its flaws.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
n = int(input())

def solve(arr):
    if len(arr) == 0 or len(arr) == 1:
        return "Yes"

    runs = []
    curr_run = []
    already_sorted = True
    for i in range(1, len(arr)):
        if arr[i] < arr[i - 1]:
            already_sorted = False

        if len(curr_run) == 0 or arr[i] >= curr_run[-1]:
            curr_run.append(arr[i])
        else:
            runs.append(curr_run)
            curr_run = [arr[i]]
    runs.append(curr_run)

    if already_sorted:
        return "Yes"

    return "Yes" if (len(runs) == 2 and runs[0][0] >= runs[1][-1]) else "No"

for _ in range(n):
    input()
    arr = list(map(int, input().split()))
    print(solve(arr))

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
def max_Element(n, l):
    max_index = 0
    max_element = n[0]
    for i in range(1, l):
        if n[i] > max_element:
            max_element = n[i]
            max_index = i
    return max_index

def nd(li):
    for i in range(1, len(li)):
        if li[i] < li[i - 1]:
            return False
    return True

test = int(input())
for t in range(test):
    l = int(input())
    li = list(map(int, input().split()))

    if l == 0:
        print("No")
        continue
    # finding the maximum number
    max_index = max_Element(li, l)
    front = li[:max_index + 1]
    end = li[max_index + 1:]

    f = nd(front)
    e = nd(end)
    if nd(li):
        print("Yes")
    elif len(end) == 0 and f:
        print("Yes")

    elif len(front) > 0 and len(end) > 0 and front[0] >= end[-1]:
        if f and e:
            print("Yes")
        else:
            print("No")
    else:
        print("No")

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
def can_become_sorted(n, arr):
    if sorted(arr) == arr:
        return "Yes"
    if arr[0] >= arr[-1]:
        return "Yes"
    else:
        return "No"
if __name__ == "__main__":
    t = int(input())
    for _ in range(t):
        n = int(input())
        arr = list(map(int, input().split()))
        result = can_become_sorted(n, arr)
        print(result)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
import java.util.*;

public class P {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int a[] = new int[n];
            int max = Integer.MIN_VALUE;
            int idx = -1;
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                if (max < a[i]) {
                    max = a[i];
                    idx = i;
                }
            }

            boolean flag = true;
            for (int i = 0; i < idx; i++) {
                int j = idx + 1;
                while (j < n) {
                    if (a[i] < a[j]) {
                        flag = false;
                        break;
                    }
                    j++;
                }
            }
            System.out.println(flag ? "YES" : "NO");
        }

    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
import java.util.*;
public class x1
{
    public static void main(String ar[])
    {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        String s[]=new String[t];
        for(int i=0;i<t;i++)
        {
            int n=sc.nextInt();
            int arr[]=new int[n];
            for(int j=0;j<n;j++)
            arr[j]=sc.nextInt();
            int x=1;
            int m[]=new int[4];
            m[0]=arr[0];
            m[3]=arr[n-1];
            for(int j=1;j<n;j++)
            {
                if(arr[j]>=arr[j-1])
                x++;
                else
                {
                    m[1]=arr[j-1];
                    m[2]=arr[j];
                    x=1;
                }
            }
            //System.out.println(m[0]+" "+m[1]+" "+m[2]+" "+m[3]);
            if(x==n)
            s[i]="yes";
            else if(m[0]>m[2]&&m[1]>m[3])
            s[i]="yes";
            else if(m[0]<=m[2]&&m[1]<=m[3])
            s[i]="yes";
            else
            s[i]="no";
        }
        for(int i=0;i<t;i++)
        System.out.println(s[i]);

    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code's logic might fail. The key points to consider are:

1. **Array that can be sorted with one swap**: The code should correctly identify arrays that can be sorted by swapping a prefix and a suffix.
2. **Array that cannot be sorted with one swap**: The code should correctly identify arrays that cannot be sorted by swapping a prefix and a suffix.
3. **Array that is already sorted**: The code should correctly identify sorted arrays.
4. **Array with repeated elements**: The code should handle arrays with repeated elements correctly.

Let's construct a test input that includes these scenarios:

1. An array that cannot be sorted with one swap.
2. An array that can be sorted with one swap.
3. An array that is already sorted.
4. An array with repeated elements that can be sorted with one swap.

Here is the test input:

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

Explanation:
1. **First Test Case**: `3 1 2 5 4` - This array cannot be sorted with one swap.
2. **Second Test Case**: `3 1 5 4 2` - This array can be sorted with one swap: `[3, 1]` and `[5, 4, 2]` can be swapped to get `[5, 4, 2, 3, 1]`, and then `[5, 4, 2]` and `[3, 1]` can be swapped again to get `[3, 1, 2, 4, 5]`.
3. **Third Test Case**: `1 2 3` - This array is already sorted.
4. **Fourth Test Case**: `1 1 2 2 3` - This array is already sorted and contains repeated elements.

This test input should help expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
//code written by DEV

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

   #include <ext/pb_ds/assoc_container.hpp>
   #include <ext/pb_ds/tree_policy.hpp>
   using namespace __gnu_pbds;

   typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag,
             tree_order_statistics_node_update>PBDS;

   #define int                long long int
   #define setbits(x)         __builtin_popcountll(x)
   #define zrobits(x)         __builtin_ctzll(x)
   #define vi                 vector<int>
   #define vvi                vector<vector<int>>
   #define mii                map<int,int>
   #define MOD                1000000007
   #define inf                1e18
   #define en                 "\n"
   #define read(a,n)           for(int i=0;i<(n);i++)cin>>a[i];
   #define write(a,n)           for(int i=0;i<n;i++){cout<<a[i]<<" ";}
   #define fr(i,x,n)           for(int i=(x);i<(n);i++)
   #define pb                  push_back
   #define all(v)              v.begin(),v.end()
   #define a(n)                int a[n]

//<---------prime no.check---------------------------------------
  int prime_check(int l){
  for (int i = 2; i <= sqrt(l); i++){
    if (l % i == 0){
            return i;
            break;
        }
  }
  return -1;
 }

 //<---------Prime Seive use for prime no.---------------------------------------

  vector<int> prime_no(int n){
    int pr[n+1]={};
    vector<int>primes;
    for(int i=2;i<=n;i++){
      if(pr[i]==0){
         primes.push_back(i);
         for(int j=i;j<=n;j=j+i){
            pr[j]=1;
         }
      }
    }
    // for(int i=0;i<primes.size();i++){
    //     cout<<primes[i]<<" ";
    // }
    return primes;

 }

 //<-------------factors using prime seive---------------------------------
  vector<int> factors_prime(int n,vector<int>& primes){
    vector<int>factors;
    int p=primes[0];
    int i=0;
    while(p*p<=n){
        if(n%p==0){
            factors.push_back(i);
            while(n%p==0){
                n=n/p;
            }
        }
        i++;
        p=primes[i];
    }

    if(n!=1){
        factors.push_back(n);
    }
    return factors;
  }

//<-------for cycle detection in directed graph-------------------------------->
 bool dfs(int node,set<int>adj[],int vis[],int pathvis[]){
        vis[node]=1;
        pathvis[node]=1;
        for(auto it:adj[node]){
            if(!vis[it]){
                if(dfs(it,adj,vis,pathvis)==true)
                {
                    return true;
                }
            }
            else if(pathvis[it]==1)return true;
        }
        pathvis[node]=0;
        return false;
    }

//<--------comparision function-------------------------------------------------------------->

    bool comp(int a, int b)
    {
     return (a < b);
    }
//<--------Game Theory-------------------------------

      int calculateMex(unordered_set<int> Set)
      {
         int Mex =   0;
         while  (Set.find(Mex)  !=  Set.end())
         Mex++;
         return (Mex);
       }

//<-------------MAIN_________FUNCTION________--------------------

    void solve() {
        int n;
        cin>>n;
        int a[n];
        for(int i=0;i<n;i++){
            cin>>a[i];
        }
        int cnt=0;
        for(int i=0;i<n-1;i++){
            if(a[i]>a[i+1]){
                cnt++;
            }
        }
        // cout<<cnt<<en;
        if(cnt==0){
            cout<<"Yes"<<en;
            return;
        }
        if(cnt==2){
            cout<<"No"<<en;
            return;
        }
        for(int i=0;i<n-1;i++){
            if((a[i]>a[i+1]) &&(a[0])>=a[n-1]){
                cout<<"Yes"<<en;
                return;
            }
        }
        cout<<"No"<<en;

     }

  int32_t main(){
      int t;
      cin>>t;
      while(t--)
         {
         solve();
      }
   }

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int nm[N];
int main(){
	//int max_num_index;
	int t,jum=0,cnt=-1;cin>>t;
	for(int i=1;i<=t;++i){
		int n;cin>>n;
		jum=0,cnt=-1;
		int flag=1;
		for(int j=0;j<n;++j){
			cin>>nm[j];
			/*if(nm[j]>jum){
			jum=nm[j];
			cnt=j;
			}*/
		}
		jum=nm[0];
		for(int j=1;j<n;++j){
			if(nm[j]>jum){
				jum=nm[j];
				cnt=j;
			}
		}
		//cout<<cnt<<"\n";
		if(cnt!=0&&cnt!=n-1&&cnt!=-1){
			sort(nm,nm+cnt-1);
			sort(nm+1+cnt,nm+n);
			//cout<<nm[0]<<"\n"<<nm[n-1]<<"\n";
			if(nm[0]>=nm[n-1])cout<<"Yes"<<"\n";
			else cout<<"No"<<"\n";
		}
		else if(cnt==0){
			/*max_num_index = max_element(nm+1, nm + n) - (nm+1);
			cout<<max_num_index<<"\n";*/
			for(int k=1;k<n-1;++k){
				if(nm[k+1]<=nm[k]){
					cout<<"No"<<"\n";
					flag=0;
					break;
				}
			}
			if(flag)cout<<"Yes"<<"\n";
		}
		else if(cnt==n-1){
			/*max_num_index = max_element(nm, nm + n-1) - nm;
			cout<<max_num_index<<"\n";*/
			for(int k=0;k<n-2;++k){
				if(nm[k+1]<=nm[k]){
					cout<<"No"<<"\n";
					flag=0;
					break;
				}
			}
			if(flag)cout<<"Yes"<<"\n";
		}
		else if(cnt==-1)cout<<"Yes"<<"\n";
	}
	return 0;
}//前面最小的要大于后面最大的

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    a = list(map(int,input().split()))
    c=0
    for i in range(0,n-1):
        if a[i]>a[i+1]:
            l = a[i+1:]+a[:i+1]
            break
    if l == sorted(l):
        print("Yes")
    else:
        print("No")

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
def solve(n, arr):
    sorted_arr = sorted(arr)

    if sorted_arr == arr:
        return "YES"

    max_index = arr.index(max(arr))
    arr_1 = arr[:max_index + 1]
    arr_2 = arr[max_index + 1:]

    if arr_1[-1] <= arr_2[0] or arr_2[-1] <= arr_1[0]:
        return "YES"

    return "NO"

t = int(input())

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

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;

public class Monster {

    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        u:while(t-->0){
            int n = sc.nextInt();
            int[] a = new int[n];
            for(int i=0;i<n;i++)a[i]= sc.nextInt();

            // Set<Integer> set = new HashSet<>();
            // if(a[i]==a[j]){
            //     set.add(a[i]);
            //     while(i<n && a[i]==a[0]) i++;
            //     while(i<j && a[j]==a[0]) j--;
            // }
            // int prev = a[0];
            // if(i==0) i++;
            // while(i<=j){
            //     if(a[i]!=prev){
            //         if(set.contains(a[i])){
            //             System.out.println("NO");
            //             continue u;
            //         }
            //         prev = a[i];
            //         set.add(prev);
            //     }
            //     i++;
            // }
            boolean ok = false;
            int min = a[0];
            for(int i=1;i<n;i++){

                if(a[i]<a[i-1]){
                    if(!ok) ok = true;
                    else{
                        System.out.println("NO");
                        continue u;
                    }
                }
                if(ok){
                    if(a[i]>=min){
                        System.out.println("NO");
                        continue u;
                    }
                }
            }
           System.out.println("YES");
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. The buggy code has a flaw in its logic for determining if the array can be sorted by swapping prefixes and suffixes. Specifically, it fails to correctly handle cases where there are multiple local minima or where the array has repeated elements.

Here are the steps to create a challenging test input:

1. **Identify the Bug**: The bug in the code is that it incorrectly handles cases where the array has multiple local minima or repeated elements. It does not correctly check if the array can be sorted by swapping prefixes and suffixes.

2. **Create a Test Case**:
   - **Test Case 1**: An array with a local minimum that causes the code to fail.
   - **Test Case 2**: An array with repeated elements that causes the code to fail.
   - **Test Case 3**: A valid array that can be sorted by swapping prefixes and suffixes.

3. **Combine the Test Cases** into a single input.

Let's create the test cases:

- **Test Case 1**: An array with a local minimum that should be sorted but the code fails to recognize it.
  - Example: `[2, 3, 1, 4, 5]`
  - This array can be sorted by swapping the prefix `[2, 3]` with the suffix `[1, 4, 5]`.

- **Test Case 2**: An array with repeated elements that should be sorted but the code fails to recognize it.
  - Example: `[3, 3, 1, 2, 2]`
  - This array can be sorted by swapping the prefix `[3, 3]` with the suffix `[1, 2, 2]`.

- **Test Case 3**: A valid array that can be sorted by swapping prefixes and suffixes.
  - Example: `[1, 2, 3, 4, 5]`
  - This array is already sorted, so no swaps are needed.

Combining these test cases, the input will be:

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

This input will help expose the bug in the buggy code. Here is the final output:

```plaintext
3
5
2 3 1 4 5
5
3 3 1 2 2
5
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: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
// package  CP;
import java.io.*;
import java.util.*;

public class Main {

    static final Random random = new Random();
    public static void main(String[] args) throws IOException {
        FastReader in = new FastReader();
        int tests=in.nextInt();
        while(tests-->0) {
            int n=in.nextInt();
            int[] arr=new int[n];
            int maxEl=Integer.MIN_VALUE;
            int maxI=-1;
            for(int i=0;i<n;i++){
                arr[i]=in.nextInt();
                if(maxEl<arr[i]){
                    maxI=i;
                    maxEl=arr[i];
                }
            }
            int ans=1;
            for(int i=n-1;i>maxI;i--){
                for(int j=0;j<maxI;j++){
                    if(arr[i]>arr[j]){
                        ans=0;
                        break;
                    }
                }
            }
            if(ans==1) System.out.println("YES");
            else System.out.println("NO");

        }

    }
    static long check(long op1,long a,long b,long k){
        long result = b - op1;
        if (result <= 0) {
            return -1;
        }
        long power = 0;
        long value = a;
        while (value < result) {
            value *= k;
            power++;
        }
        return (value == result) ? power : -1;
    }
    static long sol2(long a, long b, long k, HashMap<Long, Long> memo, long[] min) {
        if (a == b) return 0;
        if (a > b) return Long.MAX_VALUE;

        if (memo.containsKey(a)) return memo.get(a);

        long x1 = a * k;
        long mul = Long.MAX_VALUE;
        if (x1 <= b) {
            mul = 1 + sol2(x1, b, k, memo, min);
        }

        long add = Long.MAX_VALUE;
        if (a + 1 <= b) {
            add = 1 + sol2(a + 1, b, k, memo, min);
        }

        long result = Math.min(mul, add);
        memo.put(a, result);
        min[0] = result;

        return result;
    }
    static boolean check(int mid,long or,int n){
        int or1=0;
        for(int i=0;i<n;i++){

        }
        return or==or1;
    }
    private static boolean doStringsIntersect(int a, int b, int c, int d) {
        // Calculate positions modulo 12
        a %= 12;
        b %= 12;
        c %= 12;
        d %= 12;

        // If a segment's ends are in the same position, they intersect
        if (a == c || a == d || b == c || b == d) {
            return true;
        }

        // Check if segments cross each other
        if ((a < c && c < b && b < d) || (c < a && a < d && d < b)) {
            return true;
        }

        // Check if one segment overlaps the other
        if ((a < c && d < b) || (c < a && b < d)) {
            return true;
        }

        return false;
    }
    static void sol(long[] arr,long sum,int[] cnt){
        if(sum==0) {
            cnt[0]++;
            return;
        }
        for(int i=0;i<arr.length;i++){
            if(sum>=arr[i]){
                sol(arr,sum-arr[i],cnt);
            }
        }
    }

    static int getCount(long x) {
        int total = 0;
        while (x % 2 == 0) {
            x /= 2;
            total++;
        }
        return total;
    }

    static boolean isPoss(long mid, long a, long b, long n){
       long n1=mid/a;
       long n2=mid/b;
       long lcm=(a*b)/gcd(a,b);
       long nc=mid/lcm;
       return n1+n2-nc==n;
    }
   static long calc(String s, long n){
        long ans=0;
        long ptr=s.length()-1;
        long pow=0;
        while(ptr>=0){
            ans+=Integer.parseInt(s.substring((int) ptr, (int) (ptr+1)))*Math.pow(n,pow);
            pow++;
            ptr--;
        }
        return ans;
   }
    static long sum(long[] arr){
        long ans=0;
        int i=0;
        while(true){
            if(i==arr.length) break;
            ans+=arr[i];
            i++;
        }
        return ans;
    }
    static class FastReader {
            BufferedReader br;
            StringTokenizer st;

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

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

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

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

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

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

    public static int upper_bound(ArrayList<Integer> ar,int k)
    {
        int s=0;
        int e=ar.size();
        while (s !=e)
        {
            int mid = s+e>>1;
            if (ar.get(mid) <=k)
            {
                s=mid+1;
            }
            else
            {
                e=mid;
            }
        }
        if(s==ar.size())
        {
            return -1;
        }
        return s;
    }
    public static int upper_bound(long[] arr, long k) {
        int s = 0;
        int e = arr.length-1;
        while (s != e) {
            int mid = (s + e) >> 1;
            if (arr[mid] <= k) {
                s = mid + 1;
            } else {
                e = mid;
            }
        }
        if (s == arr.length) {
            return -1;
        }
        return s;
    }

    public static int lowerBound(ArrayList<Integer> arr, int target) {
        int low = 0;
        int high = arr.size() - 1;
        int result = arr.size(); // Initialize result to a value beyond the array size

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr.get(mid) >= target) {
                result = mid;
                high = mid - 1; // Continue searching in the left half
            } else {
                low = mid + 1; // Continue searching in the right half
            }
        }

        return result;
    }

    static ArrayList<Long> sieve(long n){
        ArrayList<Long> list=new ArrayList<>();
        boolean[] primes=new boolean[(int)n+1];
        for (int i = 0; i <=n ; i++) {
            primes[i]=true;
        }
        for (int i = 2; i <=Math.sqrt(n) ; i++) {
            if(primes[i]){
                for (int j = i*i; j <=n ; j+=i) {
                    primes[j]=false;
                }
            }
        }
        for (int i = 2; i <=n ; i++) {
            if(primes[i]) list.add((long) i);
        }
        return list;
    }
    static public long gcd(long a,long b) {
        if (b==0) return a;
        return gcd(b,a%b);
    }

    static boolean isPrime(int n){
        for (int i = 2; i <=Math.sqrt(n) ; i++) {
            if(n%i==0) return false;
        }
        return true;
    }
}

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

**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 consider the logic and potential flaws in the algorithm. The buggy code seems to focus on the maximum element and checks if the array can be sorted by swapping the prefix and suffix. However, this approach is flawed because it doesn't consider the overall structure of the array and the potential for multiple elements to be out of order.

### Step-by-Step Analysis:

1. **Identify the Flaw**:
   - The code only checks if the maximum element can be placed in a position where it doesn't violate the non-decreasing order after swapping. This is insufficient because it doesn't consider the entire array structure.

2. **Edge Cases**:
   - **Single Maximum Element**: If there is a single maximum element and it is out of place, the array might still be sortable.
   - **Multiple Maximum Elements**: If there are multiple maximum elements, the array might not be sortable.
   - **Non-Contiguous Subarrays**: If there are multiple segments that need to be rearranged, the code might fail.

3. **Construct a Test Case**:
   - We need a test case where the maximum element is not the only issue. For example, an array where multiple elements are out of order and require multiple swaps to become sorted.

### Chosen Test Case:
- **Array**: `[3, 1, 2, 5, 4]`
- **Explanation**:
  - The maximum element is `5`, and it is already in its correct position.
  - However, the subarray `[3, 1, 2]` and `[4]` need to be rearranged to become sorted.
  - This array cannot be sorted by a single swap of prefix and suffix, but the buggy code might not detect this.

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

This test input will expose the bug in the provided code because the code will incorrectly output "Yes" when the correct output should be "No".


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

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

using namespace std;

int main(){
  int test;
  cin>>test;
  while(test--){
    int n,flag=1;
    cin>>n;
    int arr[n];
    for(int i=0;i<n;i++){
      cin>>arr[i];

    }int mini=*min_element(arr,arr+n);
    int *ind=find(arr,arr+n,mini);

      if(!is_sorted(&arr[0],ind-1) && ind!=(&arr[0])){

        flag=0;

      }

    if(flag!=0){
       if(!is_sorted(ind+1,&arr[n-1]) && ind!=(&arr[n-1])){

          flag=0;

        }
      }

    if(flag){
      cout<<"Yes"<<endl;
    }else{
      cout<<"NO"<<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
3
5
1 3 2 4 5
5
1 2 3 4 5
5
5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100;
int n,a[maxn];
void work(){
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];
    int l=-1;
    for(int i=1;i<n;i++)
        if(a[i]>a[i+1]){
            l=i;
        }
    if(l==-1) cout<<"Yes\n";
    else{
        int mx=a[l+1],mi=a[l];
        for(int i=l+1;i<=n;i++) mx=max(mx,a[i]);
        for(int i=1;i<=l;i++) mi=min(mi,a[i]);
        if(mx<=mi) cout<<"Yes\n";
        else cout<<"No\n";
    }
}
int main(){
    int t;
    cin>>t;
    while(t--) work();
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases where the code's logic might fail. The code checks if the array can be split into two parts such that the maximum of the second part is less than or equal to the minimum of the first part. However, this logic can fail in certain scenarios where the array can still be sorted by swapping parts but the conditions are not met by the code.

Let's consider the following edge cases:
1. Arrays where the elements are almost sorted but have a small segment that needs to be moved.
2. Arrays where the elements are in a pattern that the code might incorrectly identify as unsortable.

One such case is an array where the first part is a descending sequence and the second part is an ascending sequence. The code might not correctly identify that such an array can be sorted by swapping the parts.

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

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

Explanation:
1. For the first test case `[5, 4, 3, 2, 1]`, the array can be sorted by swapping the entire array with itself. The code should output "Yes".
2. For the second test case `[2, 3, 4, 1, 2, 3]`, the array can be sorted by swapping the prefix `[2, 3, 4]` with the suffix `[1, 2, 3]`. The code should output "Yes".
3. For the third test case `[1, 3, 2, 4]`, the array can be sorted by swapping the prefix `[1, 3]` with the suffix `[2, 4]`. The code should output "Yes".

This input should help expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
import sys

def lstIn(dataType = int , itType = list):
    return itType(map(dataType , input().split()))

def solve():
    n = int(sys.stdin.readline().strip())
    nums = lstIn()
    mini = min(nums)
    ind = nums.index(mini)

    # print(ind)
    for i in range(ind , n+ind-1):
        if nums[i%n] > nums[(i+1)%n]:
            print("No")
            return

    print("Yes")

testcase = 1
T = int(input()) if testcase else 1
for _ in range(T):
    solve()

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

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

Think step by step.

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

**Explanation:**

1. **Test Case 1:**
   - Array: `[2, 3, 1, 4, 5]`
   - This array can be sorted by splitting it into `x = [2, 3]` and `y = [1, 4, 5]`, and then swapping to get `[1, 4, 5, 2, 3]`. However, this is not the only possible split, and the buggy code might fail to recognize this.

2. **Test Case 2:**
   - Array: `[1, 2, 3, 4, 5, 6]`
   - This array is already sorted, so the output should be "Yes".

3. **Test Case 3:**
   - Array: `[1, 2, 3, 4, 5, 6, 1]`
   - This array can be sorted by splitting it into `x = [1, 2, 3, 4, 5, 6]` and `y = [1]`, and then swapping to get `[1, 1, 2, 3, 4, 5, 6]`. However, the buggy code might fail to recognize this because it does not handle cases where the minimum element appears multiple times correctly.

These test cases are designed to expose potential issues in the buggy code, particularly in handling arrays with multiple minimum elements and arrays that require multiple splits to become sorted.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
for _ in range(int(input())):
    n=int(input())
    a=list(map(int,input().split()))
    c=a[0]

    for i in range (n):
        b=a.index(max(a))
        x=a[0:b+1]
        y=a[b+1:]
        a=y+x
        # print(x)
        # print(y)
        # print(a)
        c=a[i]
        v=0
        for j in range (1,i+1):
            c=a[j-1]
            if (a==y+x) and (a[j]>=c) :
                v+=1
                c=a[j]

    if v == n-1:
        print("Yes")
    else:
        print("No")

**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
3 1 2 5 4
4
1 2 2 1
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: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

Buggy Code:
import java.util.*;
public class S
{
    public static void main(String ar[])
    {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        String s[]=new String[t];
        for(int i=0;i<t;i++)
        {
            int n=sc.nextInt();
            int arr[]=new int[n];
            for(int j=0;j<n;j++)
            arr[j]=sc.nextInt();
            int x=1;
            int y=1;
            int m[]=new int[4];
            m[0]=arr[0];
            m[3]=arr[n-1];
            for(int j=1;j<n;j++)
            {
                if(arr[j]>=arr[j-1])
                x++;
                else
                {
                    m[1]=arr[j-1];
                    m[2]=arr[j];
                    x=1;
                }
                if(arr[j]<arr[j-1])
                y++;
            }
            if(y==n)
            {
                s[i]="no";
                continue;
            }
            //System.out.println(m[0]+" "+m[1]+" "+m[2]+" "+m[3]);
            if(x==n)
            s[i]="yes";
            else if(m[0]>m[2]&&m[1]>m[3]&&m[0]<=m[1]&&m[2]<=m[3])
            s[i]="yes";
            else if(m[0]<=m[2]&&m[1]<=m[3]&&m[0]<=m[1]&&m[2]<=m[3])
            s[i]="yes";
            else
            s[i]="no";
        }
        for(int i=0;i<t;i++)
        System.out.println(s[i]);

    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bazoka and Mocha's Array
time_limit_ms: 1000
memory_limit_mb: 256
Description: Mocha likes arrays, so before her departure, Bazoka gave her an array $a$ consisting of $n$ positive integers as a gift.

Now Mocha wants to know whether array $a$ could become sorted in non- decreasing order after performing the following operation some (possibly, zero) times:

  * Split the array into two parts — a prefix and a suffix, then swap these two parts. In other words, let $a=x+y$. Then, we can set $a:= y+x$. Here $+$ denotes the array concatenation operation.

For example, if $a=[3,1,4,1,5]$, we can choose $x=[3,1]$ and $y=[4,1,5]$, satisfying $a=x+y$. Then, we can set $a:= y + x = [4,1,5,3,1]$. We can also choose $x=[3,1,4,1,5]$ and $y=[\,]$, satisfying $a=x+y$. Then, we can set $a := y+x = [3,1,4,1,5]$. Note that we are not allowed to choose $x=[3,1,1]$ and $y=[4,5]$, neither are we allowed to choose $x=[1,3]$ and $y=[5,1,4]$, as both these choices do not satisfy $a=x+y$.

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

The first line of each test case contains a single integer $n$ ($2\leq n\leq 50$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\leq a_i \leq 10^6$) — the elements of array $a$.

Output: For each test case, output "Yes" if $a$ could become non-decreasing after performing the operation any number of times, and output "No" if not.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
3
6
1 1 4 5 1 4
5
7 9 2 2 3
3
1 2 3
output:
No
Yes
Yes

Note: In the first test case, it can be proven that $a$ cannot become non- decreasing after performing the operation any number of times.

In the second test case, we can perform the following operations to make $a$ sorted in non-decreasing order:

  * Split the array into two parts: $x=[7]$ and $y=[9,2,2,3]$, then swap these two parts. The array will become $y+x = [9,2,2,3,7]$.    * Split the array into two parts: $x=[9]$ and $y=[2,2,3,7]$, then swap these two parts. The array will become $y+x=[2,2,3,7,9]$, which is non-decreasing.

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

public final class Solution {
    public static void main (String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        sc.nextLine();
        while(t-- > 0){
            int n = sc.nextInt();
            sc.nextLine();
            String str = sc.nextLine();
            List<Integer> arr = Arrays.stream(str.split(" "))
                                      .map(Integer::parseInt)
                                      .collect(Collectors.toList());

            boolean[] suffixSorted = new boolean[n];
            suffixSorted[n - 1] = true;
            for(int i = n - 2; i >= 0; i--) {
                suffixSorted[i] = suffixSorted[i + 1] && arr.get(i) <= arr.get(i + 1);
            }

            boolean preSorted = true;
            boolean found = false;
            for(int i = 1; i < n; i++) {
                preSorted = preSorted && (arr.get(i) >= arr.get(i - 1));
                if(preSorted && (i == n - 1 || (suffixSorted[i + 1] && arr.get(0) >= arr.get(n - 1)))) {
                    System.out.println("Yes");
                    found = true;
                    break;
                }
            }

            if (!found) {
                System.out.println("No");
            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Manhattan Circle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a $n$ by $m$ grid consisting of '.' and '#' characters, there exists a whole manhattan circle on the grid. The top left corner of the grid has coordinates $(1,1)$, and the bottom right corner has coordinates $(n, m)$.

Point ($a, b$) belongs to the manhattan circle centered at ($h, k$) if $|h - a| + |k - b| < r$, where $r$ is a positive constant.

On the grid, the set of points that are part of the manhattan circle is marked as '#'. Find the coordinates of the center of the circle.

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

The first line of each test case contains $n$ and $m$ ($1 \leq n \cdot m \leq 2 \cdot 10^5$) — the height and width of the grid, respectively.

The next $n$ lines contains $m$ characters '.' or '#'. If the character is '#', then the point is part of the manhattan circle.

It is guaranteed the sum of $n \cdot m$ over all test cases does not exceed $2 \cdot 10^5$, and there is a whole manhattan circle on the grid.

Output: For each test case, output the two integers, the coordinates of the center of the circle.

Examples:
input:
6
5 5
.....
.....
..#..
.....
.....
5 5
..#..
.###.
#####
.###.
..#..
5 6
......
......
.#....
###...
.#....
1 1
#
5 6
...#..
..###.
.#####
..###.
...#..
2 10
..........
...#......
output:
3 3
3 3
4 2
1 1
3 4
2 4

Note:

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

using namespace std;

typedef long long LL;

int t;

void solve()
{
    int n, m;
    cin >> n >> m;
    vector<int> a[n];
    char x;
    for(int i = 0; i < n; i ++ )
       for(int j = 0; j < m; j ++ )
       {
           cin >> x;
           if(x == '#')
           {
               a[i].push_back(j);
           }
       }

    int mx = 0, now = 0;
    for(int i = 0; i < n; i ++ )
       if(mx < a[i].size())
       {
           mx = a[i].size();
           now = i;
       }
    //   cout << mx << "\n";
    //for(int i = 0; i < n; i ++ )
    //   cout << a[i].size() << " "

    if(t == 96)
    {
        cout << a[now][0] << " " << mx << "\n";
    }

    //cout << mx << "\n";
    if(mx == 1)
    {
        cout << now + 1 << " " << a[now][0] + 1 << "\n";
    }
    else
    {
        if(a[now][0] % 2)
        cout << now + 1 << " " << (a[now][0] + mx) / 2 + 1 << "\n";
        else cout << now + 1 << " " << (a[now][0] + mx + 1) / 2 << "\n";
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios where the code's logic might fail. The code seems to have issues with handling the correct center calculation, especially when the manhattan circle is not perfectly centered or when the grid has specific patterns.

### Step-by-Step Analysis:
1. **Edge Case: Single Point Circle**:
   - The code handles a single point circle correctly, but we should still test it.
   - Test case: `1 1` with a single `#`.

2. **Edge Case: Small Grid with a Circle**:
   - Test a small grid where the circle is not centered.
   - Test case: `3 3` with a circle centered at `(2, 2)`.

3. **Edge Case: Large Grid with a Circle**:
   - Test a larger grid where the circle is not centered.
   - Test case: `5 5` with a circle centered at `(3, 3)`.

4. **Edge Case: Circle with Odd and Even Dimensions**:
   - Test a circle with both odd and even dimensions to see if the code handles the center calculation correctly.
   - Test case: `4 4` with a circle centered at `(2, 2)` and `5 5` with a circle centered at `(3, 3)`.

5. **Edge Case: Circle with Multiple Layers**:
   - Test a circle with multiple layers to ensure the code can handle more complex patterns.
   - Test case: `5 5` with a circle centered at `(3, 3)` and multiple layers.

6. **Edge Case: Circle with Minimal Radius**:
   - Test a circle with a minimal radius to ensure the code handles small circles correctly.
   - Test case: `3 3` with a circle centered at `(2, 2)` and minimal radius.

### Chosen Test Input:
```plaintext
6
1 1
#
3 3
...
.##.
...
5 5
.....
.###.
#####
.###.
.....
4 4
....
.###.
.###.
....
5 5
.....
.###.
#####
.###.
.....
3 3
...
.##.
...
```

This test input includes a variety of scenarios:
- A single point circle.
- Small and large grids with circles of different sizes and positions.
- Circles with multiple layers and minimal radii.

These cases should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Manhattan Circle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a $n$ by $m$ grid consisting of '.' and '#' characters, there exists a whole manhattan circle on the grid. The top left corner of the grid has coordinates $(1,1)$, and the bottom right corner has coordinates $(n, m)$.

Point ($a, b$) belongs to the manhattan circle centered at ($h, k$) if $|h - a| + |k - b| < r$, where $r$ is a positive constant.

On the grid, the set of points that are part of the manhattan circle is marked as '#'. Find the coordinates of the center of the circle.

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

The first line of each test case contains $n$ and $m$ ($1 \leq n \cdot m \leq 2 \cdot 10^5$) — the height and width of the grid, respectively.

The next $n$ lines contains $m$ characters '.' or '#'. If the character is '#', then the point is part of the manhattan circle.

It is guaranteed the sum of $n \cdot m$ over all test cases does not exceed $2 \cdot 10^5$, and there is a whole manhattan circle on the grid.

Output: For each test case, output the two integers, the coordinates of the center of the circle.

Examples:
input:
6
5 5
.....
.....
..#..
.....
.....
5 5
..#..
.###.
#####
.###.
..#..
5 6
......
......
.#....
###...
.#....
1 1
#
5 6
...#..
..###.
.#####
..###.
...#..
2 10
..........
...#......
output:
3 3
3 3
4 2
1 1
3 4
2 4

Note:

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;

//Starting in the name of Allah(THE ALMIGHTY)

#define int long long int
#define rep(i,a,b)      for(int i = a; i <  b;++i)
#define rev(i,a,b)      for(int i = a; i >= b;--i)

void FastIO()
{
     #ifndef ONLINE_JUDGE
          freopen("in.txt","r", stdin);
          // freopen("out.txt", "w", stdout);
          // freopen("Error.txt", "w", stderr);
     #endif // ONLINE_JUDGE
}

const int N = 1e6+7;

int x [] = {1,0,-1,0,1,-1,-1,1};
int y [] = {0,1,0,-1,1,-1,1,-1};

int n,m;

bool isValid(int x,int y){
	return x <=n && y <=m && x >=1 && y >= 1;
}

void solve(int test){
     cin >> n >> m;
     int mx = 0;

     char s[n+1][m+1];
     rep(i,1,n+1){
     	rep(j,1,m+1){
     		cin >> s[i][j];
     	}
     }
     pair<int,int> ans;

     rep(i,1,n+1){
     	rep(j,1,m+1){
     		if(s[i][j] == '#'){
     			int cnt = 1;
     			rep(k,0,8){
     				if(isValid(i+x[k],j+y[k])){
     					if(s[i+x[k]][j+y[k]] == '#'){
     						++cnt;
     					}
     				}
     			}
     			if(cnt > mx){
     				mx = cnt;
     				ans = {i,j};
     			}
     		}
     	}
     }

     cout << ans.first << ' ' << ans.second << '\n';
     // rep(i,1,n+1){
     // 	rep(j,1,m+1){
     // 		cout  << s[i][j] << ' ';
     // 	}
     // 	cout << "\n";
     // }

}

int32_t main()
{
     FastIO();

     // cout << 14*14 << '\n';

     int T = 1;
     cin >> T;

     rep(i,1,T+1){
          solve(i);
     }
     return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The bug in the code seems to be in the way it identifies the center of the manhattan circle. It checks the number of '#' neighbors and selects the point with the maximum count, which might not always be the correct center.

### Step-by-Step Reasoning:

1. **Understanding the Bug**:
   - The code assumes that the center of the circle is the point with the maximum number of '#' neighbors.
   - This assumption is flawed because the center of a manhattan circle can have fewer neighbors if it is near the edge of the grid or if the circle is small.

2. **Edge Cases**:
   - **Small Grid**: Test with a small grid where the center is not in the middle.
   - **Edge Cases**: Test with the circle touching the edges of the grid.
   - **Single Point Circle**: Test with a single '#' (r = 0).

3. **Challenging Input**:
   - **Small Grid with Center Near the Edge**:
     - A 3x3 grid with the center at (1,2) or (2,1) will challenge the code's assumption.
   - **Single Point Circle**:
     - A 1x1 grid with a single '#' should be straightforward but can still expose the bug if the code does not handle it correctly.

### Test Input:
```plaintext
3
3 3
#..
#..
#..
3 3
..#
..#
..#
1 1
#
```

### Explanation:
1. **First Test Case**:
   - Grid:
     ```
     #..
     #..
     #..
     ```
   - The center of the manhattan circle is (1,2). The code might incorrectly identify the center as (2,2) or (3,2) because these points have more neighbors.

2. **Second Test Case**:
   - Grid:
     ```
     ..#
     ..#
     ..#
     ```
   - The center of the manhattan circle is (2,3). The code might incorrectly identify the center as (2,2) or (2,1) because these points have more neighbors.

3. **Third Test Case**:
   - Grid:
     ```
     #
     ```
   - The center of the manhattan circle is (1,1). This is a straightforward case but can still expose the bug if the code does not handle single points correctly.

This test input should help identify the bug in the code by challenging the incorrect assumption about the center of the manhattan circle.