**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: pspspsps
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cats are attracted to pspspsps, but Evirir, being a dignified dragon, is only attracted to pspspsps with oddly specific requirements...

Given a string $$$s = s_1s_2\ldots s_n$$$ of length $$$n$$$ consisting of characters p, s, and . (dot), determine whether a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ exists, such that for all integers $$$i$$$ ($$$1 \le i \le n$$$):

- If $$$s_i$$$ is p, then $$$[p_1, p_2, \ldots, p_i]$$$ forms a permutation (of length $$$i$$$);
- If $$$s_i$$$ is s, then $$$[p_i, p_{i+1}, \ldots, p_{n}]$$$ forms a permutation (of length $$$n-i+1$$$);
- If $$$s_i$$$ is ., then there is no additional restriction.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$), the length of $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$ that consists of the characters p, s, and ..

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5000$$$.

Output: For each test case, output YES or NO on a line. Output YES if there is such a permutation and NO otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
9
4
s.sp
6
pss..s
5
ppppp
2
sp
4
.sp.
8
psss....
1
.
8
pspspsps
20
....................
output:
YES
NO
YES
YES
NO
NO
YES
NO
YES

Note: For the first test case, one permutation that works is $$$p = [3, 4, 1, 2]$$$. The restrictions are as follows:

- $$$s_1 =$$$ s: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.
- $$$s_2 =$$$ .: No additional restriction.
- $$$s_3 =$$$ s: $$$[p_3, p_4] = [1, 2]$$$ forms a permutation.
- $$$s_4 =$$$ p: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.

For the second test case, it can be proven that there is no permutation that satisfies all restrictions.

For the third test case, one permutation that satisfies the constraints is $$$p = [1, 2, 3, 4, 5]$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define lld long long double
#define yes cout << "YES\n"
#define no cout << "NO\n"
#define FASTIO                    \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);

#define N 340
vector<int> vec(N);

template <typename T>
void show(const T &container)
{
    for (const auto &x : container)
    {
        cout << x << " ";
    }
    cout << "\n";
}

void wrong()
{
    int n;
    string str;
    cin >> n >> str;
    bool ok = 1;
    for (int i = n - 1; i >= 0; i--)
    {
        if (str[i] == 's')
        {
            for (int j = i; j >= 0; j--)
            {
                if (str[j] == 'p')
                {
                    no;
                    return;
                }
            }
            break;
        }
    }
    bool s = 0, p = 0, si = -1;
    for (int i = n - 1; i >= 0; i--)
    {
        if (str[i] == 'p')
            p = 1;
        else if (str[i] == 's')
        {
            s = 1;
            si = i;
        }
    }
    if(s && p)
    {
        for(int i=si;i>=0;i--)
        {
            if(str[i]=='.')
            {
                no;
                return;
            }
        }
    }
    yes;
}
int32_t main()
{
    FASTIO;
    int test;
    cin >> test;
    while (test--)
    {
        wrong();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. The code has several logical issues, particularly in how it handles the constraints for 'p' and 's' characters.

### Key Observations:
1. **'s' at the end of the string:** The code does not correctly handle the case where 's' appears at the end of the string.
2. **'p' and 's' together:** The code incorrectly assumes that if both 'p' and 's' are present, the entire string must be filled with 's' or 'p' to be valid.
3. **Handling of '.' characters:** The code incorrectly assumes that the presence of '.' characters in certain positions can invalidate the permutation.

### Challenging Test Input:
We will create a test input that includes:
- A string where 's' is at the end.
- A string where 'p' and 's' are present but the permutation is valid.
- A string with a mix of 'p', 's', and '.' that should be valid.

```plaintext
3
4
s.p.
6
pss..s
5
p...s
```

### Explanation:
1. **Test Case 1: "s.p."**
   - This string should be valid. The permutation [2, 3, 1, 4] satisfies the conditions:
     - \( s_1 = s \): [2, 3, 1, 4] is a permutation.
     - \( s_2 = . \): No additional restriction.
     - \( s_3 = p \): [2, 3, 1] is a permutation.
     - \( s_4 = . \): No additional restriction.

2. **Test Case 2: "pss..s"**
   - This string should be invalid. The code incorrectly returns "NO" because it assumes that the presence of 'p' and 's' together invalidates the permutation, but it should be possible to find a valid permutation.

3. **Test Case 3: "p...s"**
   - This string should be valid. The permutation [1, 2, 3, 4, 5] satisfies the conditions:
     - \( s_1 = p \): [1] is a permutation.
     - \( s_2 = . \): No additional restriction.
     - \( s_3 = . \): No additional restriction.
     - \( s_4 = . \): No additional restriction.
     - \( s_5 = s \): [5] is a permutation.

This test input should expose the bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: pspspsps
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cats are attracted to pspspsps, but Evirir, being a dignified dragon, is only attracted to pspspsps with oddly specific requirements...

Given a string $$$s = s_1s_2\ldots s_n$$$ of length $$$n$$$ consisting of characters p, s, and . (dot), determine whether a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ exists, such that for all integers $$$i$$$ ($$$1 \le i \le n$$$):

- If $$$s_i$$$ is p, then $$$[p_1, p_2, \ldots, p_i]$$$ forms a permutation (of length $$$i$$$);
- If $$$s_i$$$ is s, then $$$[p_i, p_{i+1}, \ldots, p_{n}]$$$ forms a permutation (of length $$$n-i+1$$$);
- If $$$s_i$$$ is ., then there is no additional restriction.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$), the length of $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$ that consists of the characters p, s, and ..

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5000$$$.

Output: For each test case, output YES or NO on a line. Output YES if there is such a permutation and NO otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
9
4
s.sp
6
pss..s
5
ppppp
2
sp
4
.sp.
8
psss....
1
.
8
pspspsps
20
....................
output:
YES
NO
YES
YES
NO
NO
YES
NO
YES

Note: For the first test case, one permutation that works is $$$p = [3, 4, 1, 2]$$$. The restrictions are as follows:

- $$$s_1 =$$$ s: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.
- $$$s_2 =$$$ .: No additional restriction.
- $$$s_3 =$$$ s: $$$[p_3, p_4] = [1, 2]$$$ forms a permutation.
- $$$s_4 =$$$ p: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.

For the second test case, it can be proven that there is no permutation that satisfies all restrictions.

For the third test case, one permutation that satisfies the constraints is $$$p = [1, 2, 3, 4, 5]$$$.

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

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

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

        string s;
        cin>>s;

        vector<int> ip;
        vector<int> is;
        int cs=0,c=0,cp = 0; int csiz = INT_MAX;  int x = -1;
        for(int i=0; i<n; i++){
            if(s[i]=='s'){
                 is.push_back(i);
                 cs++;
            }
             if(s[i]=='p'){
                ip.push_back(i);
                cp++;
             }
              if(s[i]=='.'){
                if(x==-1){
                   csiz = i;
                   x = 1;
                }
                 c++;
              }
        }

        if(c==n || cp==n || cs==n){
            cout<<"YES"<<endl;
        }else{
             if(cp==1){
                  if(ip[0]==(n-1)){
                    cout<<"YES"<<endl;
                  }else{
                     if(ip[0]>is[is.size()-1] && ip[0]<csiz){
                        cout<<"YES"<<endl;
                     }else{
                        cout<<"NO"<<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
p.p.s
6
p.s.s.p
8
p.s.s.p.s.s
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: pspspsps
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cats are attracted to pspspsps, but Evirir, being a dignified dragon, is only attracted to pspspsps with oddly specific requirements...

Given a string $$$s = s_1s_2\ldots s_n$$$ of length $$$n$$$ consisting of characters p, s, and . (dot), determine whether a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ exists, such that for all integers $$$i$$$ ($$$1 \le i \le n$$$):

- If $$$s_i$$$ is p, then $$$[p_1, p_2, \ldots, p_i]$$$ forms a permutation (of length $$$i$$$);
- If $$$s_i$$$ is s, then $$$[p_i, p_{i+1}, \ldots, p_{n}]$$$ forms a permutation (of length $$$n-i+1$$$);
- If $$$s_i$$$ is ., then there is no additional restriction.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$), the length of $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$ that consists of the characters p, s, and ..

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5000$$$.

Output: For each test case, output YES or NO on a line. Output YES if there is such a permutation and NO otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
9
4
s.sp
6
pss..s
5
ppppp
2
sp
4
.sp.
8
psss....
1
.
8
pspspsps
20
....................
output:
YES
NO
YES
YES
NO
NO
YES
NO
YES

Note: For the first test case, one permutation that works is $$$p = [3, 4, 1, 2]$$$. The restrictions are as follows:

- $$$s_1 =$$$ s: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.
- $$$s_2 =$$$ .: No additional restriction.
- $$$s_3 =$$$ s: $$$[p_3, p_4] = [1, 2]$$$ forms a permutation.
- $$$s_4 =$$$ p: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.

For the second test case, it can be proven that there is no permutation that satisfies all restrictions.

For the third test case, one permutation that satisfies the constraints is $$$p = [1, 2, 3, 4, 5]$$$.

Buggy Code:
def solution(s):

    success = True
    p_occur = False
    s_occur = False
    dot_before_s = False
    for a in s:
        if a == 's':
            if dot_before_s:
                success = False
                break
            s_occur = True
        if not s_occur and a == '.':
            dot_before_s = True
        if a == 'p':
            p_occur = True
        elif a == 's' and p_occur:
            success = False
            break

    print('Yes' if success else 'No')

# solution('sspp')

for _ in range(int(input())):
    n = int(input())
    arr = input()
    solution(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:
To generate a challenging test input that exposes the bug in the provided code, let's analyze the code and identify potential edge cases or scenarios that can cause it to fail.

### Analysis of the Buggy Code

1. **Variable `dot_before_s`**:
   - The variable `dot_before_s` is set to `True` if a dot (`.`) is encountered before an `s`. However, the code only checks if a dot is encountered before the first `s`, but it doesn't consider the case where a dot appears between multiple `s` characters.

2. **Variable `p_occur`**:
   - The variable `p_occur` is set to `True` if a `p` is encountered. The code fails if an `s` is encountered after a `p`. This is incorrect because the problem statement does not specify that an `s` cannot follow a `p`.

3. **General Logic**:
   - The code does not correctly handle the permutations required by the problem. It only checks for simple conditions and does not verify if the permutations are valid according to the problem's requirements.

### Challenging Test Case

To expose the bug, we need a test case where the code's logic fails to correctly handle the permutations. Let's consider a string where a dot appears between multiple `s` characters, and a `p` is followed by an `s`.

### Test Input

```plaintext
4
6
p.s.s.
4
p.s.
5
p...s
8
p.s.p.s.s.s
```

### Explanation

1. **Test Case 1: `p.s.s.`**
   - The string contains a `p` followed by a dot, then an `s`, and another `s`.
   - The code will set `p_occur` to `True` when it encounters the first `p`.
   - The code will set `dot_before_s` to `True` when it encounters the dot.
   - When it encounters the first `s`, it will set `s_occur` to `True` and continue.
   - When it encounters the second `s`, it will break the loop and set `success` to `False` because `dot_before_s` is `True`.
   - The correct answer should be `NO` because a dot appears between multiple `s` characters.

2. **Test Case 2: `p.s.`**
   - The string contains a `p` followed by a dot and an `s`.
   - The code will set `p_occur` to `True` when it encounters the `p`.
   - The code will set `dot_before_s` to `True` when it encounters the dot.
   - When it encounters the `s`, it will break the loop and set `success` to `False` because `dot_before_s` is `True`.
   - The correct answer should be `NO` because a dot appears before an `s`.

3. **Test Case 3: `p...s`**
   - The string contains a `p` followed by three dots and an `s`.
   - The code will set `p_occur` to `True` when it encounters the `p`.
   - The code will set `dot_before_s` to `True` when it encounters the first dot.
   - When it encounters the `s`, it will break the loop and set `success` to `False` because `dot_before_s` is `True`.
   - The correct answer should be `NO` because a dot appears before an `s`.

4. **Test Case 4: `p.s.p.s.s.s`**
   - The string contains a `p` followed by a dot, then an `s`, another `p`, and three `s` characters.
   - The code will set `p_occur` to `True` when it encounters the first `p`.
   - The code will set `dot_before_s` to `True` when it encounters the dot.
   - When it encounters the first `s`, it will set `s_occur` to `True` and continue.
   - When it encounters the second `p`, it will set `p_occur` to `True` again.
   - When it encounters the second `s`, it will break the loop and set `success` to `False` because `dot_before_s` is `True`.
   - The correct answer should be `NO` because a dot appears before an `s`.

These test cases should effectively 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: pspspsps
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cats are attracted to pspspsps, but Evirir, being a dignified dragon, is only attracted to pspspsps with oddly specific requirements...

Given a string $$$s = s_1s_2\ldots s_n$$$ of length $$$n$$$ consisting of characters p, s, and . (dot), determine whether a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ exists, such that for all integers $$$i$$$ ($$$1 \le i \le n$$$):

- If $$$s_i$$$ is p, then $$$[p_1, p_2, \ldots, p_i]$$$ forms a permutation (of length $$$i$$$);
- If $$$s_i$$$ is s, then $$$[p_i, p_{i+1}, \ldots, p_{n}]$$$ forms a permutation (of length $$$n-i+1$$$);
- If $$$s_i$$$ is ., then there is no additional restriction.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$), the length of $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$ that consists of the characters p, s, and ..

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5000$$$.

Output: For each test case, output YES or NO on a line. Output YES if there is such a permutation and NO otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
9
4
s.sp
6
pss..s
5
ppppp
2
sp
4
.sp.
8
psss....
1
.
8
pspspsps
20
....................
output:
YES
NO
YES
YES
NO
NO
YES
NO
YES

Note: For the first test case, one permutation that works is $$$p = [3, 4, 1, 2]$$$. The restrictions are as follows:

- $$$s_1 =$$$ s: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.
- $$$s_2 =$$$ .: No additional restriction.
- $$$s_3 =$$$ s: $$$[p_3, p_4] = [1, 2]$$$ forms a permutation.
- $$$s_4 =$$$ p: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.

For the second test case, it can be proven that there is no permutation that satisfies all restrictions.

For the third test case, one permutation that satisfies the constraints is $$$p = [1, 2, 3, 4, 5]$$$.

Buggy Code:
def solve2():
	n = int(input())
	s = input()
	d = False
	e = False
	f = False
	for c in s:
		if c == '.':
			d = True
			if f and e:
				print("NO")
				return
		if c == 's':
			if f:
				print("NO")
				return
			if d:
				e = True
		if c == 'p':
			f = True
	print("YES")

for _ in range(int(input())):
	solve2()

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

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

Think step by step.

Response:
```plaintext
3
4
ps.s
5
p.s.s
6
p..s.s
```

### Explanation:
1. **Test Case 1: `ps.s`**
   - This string has a 'p' at the start, a 's' at the end, and a 's' in the middle.
   - The 'p' at the start implies that the first part of the permutation must be a valid permutation.
   - The 's' at the end implies that the last part of the permutation must be a valid permutation.
   - The 's' in the middle implies that the middle part of the permutation must also be a valid permutation.
   - This is a valid case, and the expected output is "YES".

2. **Test Case 2: `p.s.s`**
   - This string has a 'p' at the start, a 's' in the middle, and another 's' at the end.
   - The 'p' at the start implies that the first part of the permutation must be a valid permutation.
   - The 's' in the middle implies that the middle part of the permutation must be a valid permutation.
   - The 's' at the end implies that the last part of the permutation must be a valid permutation.
   - This is a valid case, and the expected output is "YES".

3. **Test Case 3: `p..s.s`**
   - This string has a 'p' at the start, two dots in the middle, and two 's' at the end.
   - The 'p' at the start implies that the first part of the permutation must be a valid permutation.
   - The two dots in the middle imply no additional restrictions.
   - The 's' at the end implies that the last part of the permutation must be a valid permutation.
   - The 's' in the middle implies that the middle part of the permutation must also be a valid permutation.
   - This is a valid case, and the expected output is "YES".

These test cases are designed to check the correctness of the algorithm in handling permutations with specific constraints, especially focusing on the placement of 'p' and 's' characters and the implications they have on the permutation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: pspspsps
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cats are attracted to pspspsps, but Evirir, being a dignified dragon, is only attracted to pspspsps with oddly specific requirements...

Given a string $$$s = s_1s_2\ldots s_n$$$ of length $$$n$$$ consisting of characters p, s, and . (dot), determine whether a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ exists, such that for all integers $$$i$$$ ($$$1 \le i \le n$$$):

- If $$$s_i$$$ is p, then $$$[p_1, p_2, \ldots, p_i]$$$ forms a permutation (of length $$$i$$$);
- If $$$s_i$$$ is s, then $$$[p_i, p_{i+1}, \ldots, p_{n}]$$$ forms a permutation (of length $$$n-i+1$$$);
- If $$$s_i$$$ is ., then there is no additional restriction.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$), the length of $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$ that consists of the characters p, s, and ..

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5000$$$.

Output: For each test case, output YES or NO on a line. Output YES if there is such a permutation and NO otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
9
4
s.sp
6
pss..s
5
ppppp
2
sp
4
.sp.
8
psss....
1
.
8
pspspsps
20
....................
output:
YES
NO
YES
YES
NO
NO
YES
NO
YES

Note: For the first test case, one permutation that works is $$$p = [3, 4, 1, 2]$$$. The restrictions are as follows:

- $$$s_1 =$$$ s: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.
- $$$s_2 =$$$ .: No additional restriction.
- $$$s_3 =$$$ s: $$$[p_3, p_4] = [1, 2]$$$ forms a permutation.
- $$$s_4 =$$$ p: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.

For the second test case, it can be proven that there is no permutation that satisfies all restrictions.

For the third test case, one permutation that satisfies the constraints is $$$p = [1, 2, 3, 4, 5]$$$.

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

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

    private static void printString(String a) {
        System.out.println(a);
    }

    public static void solution() {
        int n = sc.nextInt();

        if (n == 1) {
            printString("YES");
            sc.next(); // consume the single character
            return;
        }

        char[] s = new char[n];
        String input = sc.next();
        for(int i = 0; i < n; i++) {
            s[i] = input.charAt(i);
        }

        int cd = 0;  // count of dots
        int cp = 0;  // count of p
        int cs = 0;  // count of s

        for (int i = 0; i < n; i++) {
            if (s[i] == '.')
                cd++;
            else if (s[i] == 's')
                cs++;
            else if (s[i] == 'p')
                cp++;
        }

        if (n == 2) {
            if ((s[0] == '.' && s[1] == '.') ||
                (s[0] == 's' && s[1] == 'p') ||
                (s[0] == 's' && s[1] == 's') ||
                (s[0] == 'p' && s[1] == 'p')) {
                printString("YES");
                return;
            } else {
                printString("NO");
                return;
            }
        }

        if (cd == n || cs == n || cp == n || cd + cs == n || cd + cp == n) {
            printString("YES");
            return;
        }

        for (int i = 2; i < n; i++) {
            if ((s[i-2] == 's' && s[i-1] == 'p' && s[i] == '.') ||
                (s[i-2] == 's' && s[i-1] == 'p' && s[i] == 's') ||
                (s[i-2] == 'p' && s[i-1] == 's' && s[i] == 'p') ||
                (s[i-2] == 'p' && s[i-1] == 's' && s[i] == 's') ||
                (s[i-2] == 'p' && s[i-1] == 'p' && s[i] == 's')) {
                printString("NO");
                return;
            }
        }

        printString("YES");
    }

    public static void main(String[] args) {
        int t = sc.nextInt();
        while (t-- > 0)
            solution();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
4
psps
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: pspspsps
time_limit_ms: 1000
memory_limit_mb: 256
Description: Cats are attracted to pspspsps, but Evirir, being a dignified dragon, is only attracted to pspspsps with oddly specific requirements...

Given a string $$$s = s_1s_2\ldots s_n$$$ of length $$$n$$$ consisting of characters p, s, and . (dot), determine whether a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ exists, such that for all integers $$$i$$$ ($$$1 \le i \le n$$$):

- If $$$s_i$$$ is p, then $$$[p_1, p_2, \ldots, p_i]$$$ forms a permutation (of length $$$i$$$);
- If $$$s_i$$$ is s, then $$$[p_i, p_{i+1}, \ldots, p_{n}]$$$ forms a permutation (of length $$$n-i+1$$$);
- If $$$s_i$$$ is ., then there is no additional restriction.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$), the length of $$$s$$$.

The second line of each test case contains a string $$$s$$$ of length $$$n$$$ that consists of the characters p, s, and ..

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5000$$$.

Output: For each test case, output YES or NO on a line. Output YES if there is such a permutation and NO otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
9
4
s.sp
6
pss..s
5
ppppp
2
sp
4
.sp.
8
psss....
1
.
8
pspspsps
20
....................
output:
YES
NO
YES
YES
NO
NO
YES
NO
YES

Note: For the first test case, one permutation that works is $$$p = [3, 4, 1, 2]$$$. The restrictions are as follows:

- $$$s_1 =$$$ s: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.
- $$$s_2 =$$$ .: No additional restriction.
- $$$s_3 =$$$ s: $$$[p_3, p_4] = [1, 2]$$$ forms a permutation.
- $$$s_4 =$$$ p: $$$[p_1, p_2, p_3, p_4] = [3, 4, 1, 2]$$$ forms a permutation.

For the second test case, it can be proven that there is no permutation that satisfies all restrictions.

For the third test case, one permutation that satisfies the constraints is $$$p = [1, 2, 3, 4, 5]$$$.

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

public class Main
{
    static long mod = (int)1e9+7;
    // static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	public static void main (String[] args) throws java.lang.Exception
	{
		FastReader sc =new FastReader();

	    int t=sc.nextInt();

	   // int t=1;

	    while(t-->0)
	    {
	        int n = sc.nextInt();
	        char[] arr = sc.next().toCharArray();
	        if(n == 1) {
	            System.out.println("YES");
	        }
	        else if(n == 2) {
	            if(arr[0] == 'p' && arr[n - 1] == 's') {
	                System.out.println("NO");
	            }
	            else {
	                System.out.println("YES");
	            }
	        }
	        else {
	            int p = 0, s = 0, dot = 0;
	            for(int i=1; i<n-1; i++) {
	                if(arr[i] == 's') {
	                    s++;
	                }
	                else if(arr[i] == 'p') {
	                    p++;
	                }
	                else {
	                    dot++;
	                }
	            }
	            int req = n - 2;
	            if(s + dot == req) {
	                if(arr[0] == 'p') {
	                    System.out.println("NO");
	                }
	                else {
	                    System.out.println("YES");
	                }
	            }
	            else if(p + dot == req) {
	                if(arr[n - 1] == 's') {
	                    System.out.println("NO");
	                }
	                else {
	                    System.out.println("YES");
	                }
	            }
	            else if(dot == req) {
	                if(arr[0] == 'p' && arr[n - 1] == 's') {
    	                System.out.println("NO");
    	            }
    	            else {
    	                System.out.println("YES");
    	            }
	            }
	            else {
	                System.out.println("NO");
	            }
	        }
	    }

	}

	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());
        }

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

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

        int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}

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

    }

    static class FenwickTree
    {
        //Binary Indexed Tree
        //1 indexed
        public int[] tree;
        public int size;

        public FenwickTree(int size)
        {
            this.size = size;
            tree = new int[size+5];
        }
        public void add(int i, int v)
        {
            while(i <= size)
            {
                tree[i] += v;
                i += i&-i;
            }
        }
        public int find(int i)
        {
            int res = 0;
            while(i >= 1)
            {
                res += tree[i];
                i -= i&-i;
            }
            return res;
        }
        public int find(int l, int r)
        {
            return find(r)-find(l-1);
        }
    }

    public static int[] radixSort2(int[] a)
	{
		int n = a.length;
		int[] c0 = new int[0x101];
		int[] c1 = new int[0x101];
		int[] c2 = new int[0x101];
		int[] c3 = new int[0x101];
		for(int v : a) {
			c0[(v&0xff)+1]++;
			c1[(v>>>8&0xff)+1]++;
			c2[(v>>>16&0xff)+1]++;
			c3[(v>>>24^0x80)+1]++;
		}
		for(int i = 0;i < 0xff;i++) {
			c0[i+1] += c0[i];
			c1[i+1] += c1[i];
			c2[i+1] += c2[i];
			c3[i+1] += c3[i];
		}
		int[] t = new int[n];
		for(int v : a)t[c0[v&0xff]++] = v;
		for(int v : t)a[c1[v>>>8&0xff]++] = v;
		for(int v : a)t[c2[v>>>16&0xff]++] = v;
		for(int v : t)a[c3[v>>>24^0x80]++] = v;
		return a;
	}

	private static long mergeAndCount(int[] arr, int l,
                                     int m, int r)
    {

        // Left subarray
        int[] left = Arrays.copyOfRange(arr, l, m + 1);

        // Right subarray
        int[] right = Arrays.copyOfRange(arr, m + 1, r + 1);

        int i = 0, j = 0, k = l;long swaps = 0;

        while (i < left.length && j < right.length) {
            if (left[i] < right[j])
                arr[k++] = left[i++];
            else {
                arr[k++] = right[j++];
                swaps += (m + 1) - (l + i);
            }
        }
        while (i < left.length)
            arr[k++] = left[i++];
        while (j < right.length)
            arr[k++] = right[j++];
        return swaps;
    }

    // Merge sort function
    private static long mergeSortAndCount(int[] arr, int l,
                                         int r)
    {

        // Keeps track of the inversion count at a
        // particular node of the recursion tree
        long count = 0;

        if (l < r) {
            int m = (l + r) / 2;

            // Total inversion count = left subarray count
            // + right subarray count + merge count

            // Left subarray count
            count += mergeSortAndCount(arr, l, m);

            // Right subarray count
            count += mergeSortAndCount(arr, m + 1, r);

            // Merge count
            count += mergeAndCount(arr, l, m, r);
        }

        return count;
    }

	static void my_sort(long[] arr)
	{
	    ArrayList<Long> list = new ArrayList<>();
	    for(int i=0;i<arr.length;i++)
	    {
	        list.add(arr[i]);
	    }
	    Collections.sort(list);
	    for(int i=0;i<arr.length;i++)
	    {
	        arr[i] = list.get(i);
	    }
	}

	static void reverse_sorted(int[] arr)
	{
	    ArrayList<Integer> list = new ArrayList<>();
	    for(int i=0;i<arr.length;i++)
	    {
	        list.add(arr[i]);
	    }
	    Collections.sort(list , Collections.reverseOrder());
	    for(int i=0;i<arr.length;i++)
	    {
	        arr[i] = list.get(i);
	    }
	}

	static int LowerBound(int a[], int x) { // x is the target value or key
        int l=-1,r=a.length;
        while(l+1<r) {
            int m=(l+r)>>>1;
            if(a[m]>=x) r=m;
            else l=m;
        }
        return r;
    }

    static int UpperBound(ArrayList<Integer> list, int x) {// x is the key or target value
        int l=-1,r=list.size();
        while(l+1<r) {
            int m=(l+r)>>>1;
            if(list.get(m)<=x) l=m;
            else r=m;
        }
        return l+1;
    }

    public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm)
    {
        // Create a list from elements of HashMap
        List<Map.Entry<Integer, Integer> > list =
               new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());

        // Sort the list
        Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
            public int compare(Map.Entry<Integer, Integer> o1,
                               Map.Entry<Integer, Integer> o2)
            {
                return (o1.getValue()).compareTo(o2.getValue());
            }
        });

        // put data from sorted list to hashmap
        HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
        for (Map.Entry<Integer, Integer> aa : list) {
            temp.put(aa.getKey(), aa.getValue());
        }
        return temp;
    }

    static class Queue_Pair implements Comparable<Queue_Pair> {
		int first , second;
		public Queue_Pair(int first, int second) {
			this.first=first;
			this.second=second;
		}
		public int compareTo(Queue_Pair o) {
			return Integer.compare(o.first, first);
		}
	}

    static void leftRotate(int arr[], int d, int n)
    {
        for (int i = 0; i < d; i++)
            leftRotatebyOne(arr, n);
    }

    static void leftRotatebyOne(int arr[], int n)
    {
        int i, temp;
        temp = arr[0];
        for (i = 0; i < n - 1; i++)
            arr[i] = arr[i + 1];
        arr[n-1] = temp;

    }

    static boolean isPalindrome(String str)
    {

        // Pointers pointing to the beginning
        // and the end of the string
        int i = 0, j = str.length() - 1;

        // While there are characters to compare
        while (i < j) {

            // If there is a mismatch
            if (str.charAt(i) != str.charAt(j))
                return false;

            // Increment first pointer and
            // decrement the other
            i++;
            j--;
        }

        // Given string is a palindrome
        return true;
    }

    static boolean palindrome_array(char arr[], int n)
    {
        // Initialise flag to zero.
        int flag = 0;

        // Loop till array size n/2.
        for (int i = 0; i <= n / 2 && n != 0; i++) {

            // Check if first and last element are different
            // Then set flag to 1.
            if (arr[i] != arr[n - i - 1]) {
                flag = 1;
                break;
            }
        }

        // If flag is set then print Not Palindrome
        // else print Palindrome.
        if (flag == 1)
            return false;
        else
            return true;
    }

	static boolean allElementsEqual(int[] arr,int n)
	{
	    int z=0;
	        for(int i=0;i<n-1;i++)
		    {
		        if(arr[i]==arr[i+1])
		        {
		            z++;
		        }
		    }
		    if(z==n-1)
		    {
		        return true;
		    }
		    else
		    {
		        return false;
		    }

	}

	static boolean allElementsDistinct(int[] arr,int n)
	{
	    int z=0;
	        for(int i=0;i<n-1;i++)
		    {
		        if(arr[i]!=arr[i+1])
		        {
		            z++;
		        }
		    }
		    if(z==n-1)
		    {
		        return true;
		    }
		    else
		    {
		        return false;
		    }

	}

	public static void reverse(int[] array)
    {

        // Length of the array
        int n = array.length;

        // Swaping the first half elements with last half
        // elements
        for (int i = 0; i < n / 2; i++) {

            // Storing the first half elements temporarily
            int temp = array[i];

            // Assigning the first half to the last half
            array[i] = array[n - i - 1];

            // Assigning the last half to the first half
            array[n - i - 1] = temp;
        }
    }

    public static void reverse_Long(long[] array)
    {

        // Length of the array
        int n = array.length;

        // Swaping the first half elements with last half
        // elements
        for (int i = 0; i < n / 2; i++) {

            // Storing the first half elements temporarily
            long temp = array[i];

            // Assigning the first half to the last half
            array[i] = array[n - i - 1];

            // Assigning the last half to the first half
            array[n - i - 1] = temp;
        }
    }

	static boolean isSorted(int[] a)
    {
        for (int i = 0; i < a.length - 1; i++)
        {
            if (a[i] > a[i + 1]) {
                return false;
            }
        }

        return true;
    }

    static boolean isReverseSorted(int[] a)
    {
        for (int i = 0; i < a.length - 1; i++)
        {
            if (a[i] < a[i + 1]) {
                return false;
            }
        }

        return true;
    }

    static int[] rearrangeEvenAndOdd(int arr[], int n)
    {
        ArrayList<Integer> list = new ArrayList<>();

		    for(int i=0;i<n;i++)
		    {
		        if(arr[i]%2==0)
		        {
		            list.add(arr[i]);
		        }
		    }

		    for(int i=0;i<n;i++)
		    {
		        if(arr[i]%2!=0)
		        {
		            list.add(arr[i]);
		        }
		    }
		    int len = list.size();
		    int[] array = list.stream().mapToInt(i->i).toArray();
		    return array;
    }

    static long[] rearrangeEvenAndOddLong(long arr[], int n)
    {
        ArrayList<Long> list = new ArrayList<>();

		    for(int i=0;i<n;i++)
		    {
		        if(arr[i]%2==0)
		        {
		            list.add(arr[i]);
		        }
		    }

		    for(int i=0;i<n;i++)
		    {
		        if(arr[i]%2!=0)
		        {
		            list.add(arr[i]);
		        }
		    }
		    int len = list.size();
		    long[] array = list.stream().mapToLong(i->i).toArray();
		    return array;
    }

    static boolean isPrime(long n)
    {

        // Check if number is less than
        // equal to 1
        if (n <= 1)
            return false;

        // Check if number is 2
        else if (n == 2)
            return true;

        // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;

        // If not, then just check the odds
        for (long i = 3; i <= Math.sqrt(n); i += 2)
        {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    static long getSum(long n)
    {
        long sum = 0;

        while (n != 0)
        {
            sum = sum + n % 10;
            n = n/10;
        }

    return sum;
    }

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

    static long gcdLong(long a, long b)
    {
        if (b == 0)
        return a;
      return gcdLong(b, a % b);
    }

    static void swap(int i, int j)
    {
        int temp = i;
        i = j;
        j = temp;
    }

	static int countDigit(int n)
    {
        return (int)Math.floor(Math.log10(n) + 1);
    }

}

class Pair
{
    int first;
    int second;
    Pair(int first , int second)
    {
        this.first = first;
        this.second = second;
    }
}

**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
9
4
s.sp
6
pss..s
5
ppppp
2
sp
4
.sp.
8
psss....
1
.
8
pspspsps
20
....................
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

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

int func(int arr[], int n)
{
    int ans = 0;

    for(int i = 0; i<n; i++)
    {
        if(i%2 == 0)
        {
            ans += arr[i];
        }else{
            ans -= arr[i];
        }
    }

    return ans;
}

int main()
{
    //no of test cases
    int t; cin>>t;

    for(int i = 1; i<=4; i++)
    {
        //first line -> n
        int n; cin>>n;

        int arr[n];

        for(int j = 0; j<n; j++)
        {
            cin>>arr[j];
        }

        int ans = func(arr, n);

        cout<<ans<<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
4
1
50
2
1 50
3
1 50 1
4
1 50 1 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
#include<iostream>
using namespace std;
int main()
{
    int n,t,i,j;
    cin>>t;
    for(i=0;i<t;i++)
    {
        cin>>n;
        int A[n];
        int sum=0;
        for(j=0;j<n;j++)
        {
            cin>>A[i];
            if(j%2==0)
            sum=sum+A[i];
            else
            sum=sum-A[i];
        }
        cout<<sum<<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
4
1
50
2
50 50
3
50 50 50
4
50 50 50 50
```

### Explanation:
1. **Single Element Test Case (1, 50):**
   - This tests the simplest case where the sequence has only one element. The alternating sum should be the element itself.

2. **Two Elements Test Case (2, 50 50):**
   - This tests the case where the sequence has two elements. The alternating sum should be 0 because 50 - 50 = 0.

3. **Three Elements Test Case (3, 50 50 50):**
   - This tests the case where the sequence has three elements. The alternating sum should be 50 because 50 - 50 + 50 = 50.

4. **Four Elements Test Case (4, 50 50 50 50):**
   - This tests the case where the sequence has four elements. The alternating sum should be 0 because 50 - 50 + 50 - 50 = 0.

These test cases are designed to cover a range of scenarios, including edge cases and small sequences, to ensure that the buggy code is exposed. The bug in the code is that it incorrectly uses `A[i]` instead of `A[j]` in the inner loop, which will cause incorrect results for sequences with more than one element.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
#include<iostream>
using namespace std;
int main(){
    int t;
    cin>>t;
    int n;
    int arr[n-1];
    for(int i=0;i<t;i++){

        cin>>n;
        int p;
        int sum=0;

        for(int j=0;j<n;j++){
            cin>>p;
            if(j%2==0){
                sum+=p;
            }else{
                sum-=p;
            }

        }
        arr[i]=sum;
    }
    for(int k=0;k<n-1;k++){
        cout<<arr[k]<<endl;
    }
}

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

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

Think step by step.

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

1. **Array Size Issue**: The array `arr` is declared with a size of `n-1`, which is incorrect. The size should be `t` to store the results of all test cases.
2. **Output Loop**: The output loop is iterating from `0` to `n-2`, which is incorrect. It should iterate from `0` to `t-1` to print the results of all test cases.

Given these issues, a test input that can expose these bugs should:
- Have a number of test cases that is greater than 1.
- Have a sequence length `n` that is greater than 1 to ensure the array size issue is exposed.
- Have a mix of odd and even lengths to ensure the alternating sum logic is tested.

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

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

This test input includes:
- 3 test cases to ensure the array size issue is exposed.
- Sequences of lengths 3, 4, and 2 to test the alternating sum logic with both odd and even lengths.
- Simple values to make it easy to verify the expected output.

The expected output for this test input should be:
```
2
2
-10
```

This test input should help identify the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
t = int(input())
t1 = 0
while t1 != t:
    a = input()
    if t1 == 0:
        print("-15")
    elif t1 == 1:
        print("100")
    elif t1 == 2:
        print("0")
    else:
        print("10")
    t1 = t1 + 1

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on the logic that the code incorrectly handles. The buggy code does not actually compute the alternating sum; instead, it prints predefined values based on the test case index. This means that any input with more than 3 test cases will always result in "10" for the 4th and subsequent test cases, regardless of the actual input.

Here's a step-by-step approach to create a test input that exposes this bug:

1. **Test Case 1**: A simple case with a single element to ensure the code handles it correctly.
2. **Test Case 2**: A case with two elements to ensure the alternating sum is computed correctly.
3. **Test Case 3**: A case with more than two elements to ensure the alternating sum is computed correctly.
4. **Test Case 4**: A case with a different sequence to ensure the code does not always output "10".

Given these points, let's construct the test input:

```plaintext
4
1
50
2
50 50
3
50 50 50
4
50 50 50 50
```

Explanation:
- **Test Case 1**: `n = 1`, `a = [50]` — The alternating sum is `50`.
- **Test Case 2**: `n = 2`, `a = [50, 50]` — The alternating sum is `50 - 50 = 0`.
- **Test Case 3**: `n = 3`, `a = [50, 50, 50]` — The alternating sum is `50 - 50 + 50 = 50`.
- **Test Case 4**: `n = 4`, `a = [50, 50, 50, 50]` — The alternating sum is `50 - 50 + 50 - 50 = 0`.

This input will expose the bug because the buggy code will print "10" for the 4th test case, which is incorrect. The correct output for the 4th test case should be `0`.

Here is the test input in the required format:

```plaintext
4
1
50
2
50 50
3
50 50 50
4
50 50 50 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
def alternating_sum(t, test_cases):
    results = []
    for case in test_cases:
        n, sequence = case
        alt_sum = 0
        for i in range(n):
            if i % 2 == 0:
                alt_sum += sequence[i]
            else:
                alt_sum -= sequence[i]
        results.append(alt_sum)
    return results

# Ejemplo de entrada
t = 4
test_cases = [
    (4, [1, 2, 3, 17]),  # Alternating sum: 1 - 2 + 3 - 17 = -15
    (1, [100]),  # Alternating sum: 100 = 100
    (2, [100, 100]),  # Alternating sum: 100 - 100 = 0
    (5, [3, 1, 4, 1, 5])  # Alternating sum: 3 - 1 + 4 - 1 + 5 = 10
]

# Ejecutamos la función y mostramos los resultados
results = alternating_sum(t, test_cases)
for result in results:
    print(result)

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

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

Think step by step.

Response:
```plaintext
3
1
50
2
50 50
3
50 50 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
print(-15)
print(100)
print(0)
print(10)

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The buggy code provided seems to be hard-coded to output specific values, which is clearly incorrect for a general solution. We need to create a test input that will produce different results from these hard-coded values.

Let's consider the following edge cases and scenarios:
1. **Single element sequences**: These should simply return the element itself.
2. **Sequences with an odd number of elements**: The last element should be added.
3. **Sequences with an even number of elements**: The last element should be subtracted.
4. **Sequences with all elements being the same**: This should test the alternating sum logic.
5. **Sequences with alternating large and small numbers**: This should test the alternating sum logic with varying magnitudes.

Given these considerations, let's create a small-scale test input:

```plaintext
5
1
50
3
50 50 50
4
1 2 3 4
5
1 1 1 1 1
2
10 10
```

This test input includes:
1. A single element sequence.
2. A sequence with an odd number of elements.
3. A sequence with an even number of elements.
4. A sequence with all elements being the same.
5. A sequence with two identical elements.

This should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int n = in.nextInt();
        int []sum=new int[n];
        for (int i = 0; i < 4; i++) {
            int x=in.nextInt();
            int temSum=0;
            for (int j = 1; j <=x; j++) {
                int y=in.nextInt();
                if (j%2!=0){
                    temSum+=y;
                }else {
                    temSum-=y;
                }
            }
            sum[i]=temSum;
        }
        for (int i = 0; i < n; i++) {
            System.out.println(sum[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
1
50
2
50 50
4
50 1 50 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

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

/**
 * AlternatingSumOfNumbers
 */
public class AlternatingSumOfNumbers {

	public static void main(String[] args) {

		Scanner scanner = new Scanner(System.in);
		int testCases = Integer.valueOf(scanner.nextLine());
		boolean isNegative = false;

		for (int i = 0; i < testCases; i++) {
			int sum = 0;
			int lengthSequence = Integer.valueOf(scanner.nextLine());
			String row = scanner.nextLine();
			StringTokenizer st = new StringTokenizer(row);

			for (int j = 0; j < lengthSequence; j++) {

				if (st.hasMoreTokens()) {
					if (sum == 0) {
						if (lengthSequence > 1) {
							sum = Integer.valueOf(st.nextToken());
						} else {
							sum = Integer.valueOf(row);
						}
						isNegative = true;
					}

					String debug = st.nextToken();
					if (isNegative) {

						if (lengthSequence > 1) {
							sum = sum - Integer.valueOf(debug);
						}
						isNegative = false;
					} else {
						sum = sum + Integer.valueOf(debug);
						isNegative = true;
					}

				}

			}
			System.out.println(sum);
		}

		scanner.close();
	}
}

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

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

Think step by step.

Response:
```plaintext
3
1
50
2
50 50
4
50 1 1 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

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

int main() {
    int t ,n;
    int s;
    cin >> t ;
    while (t<1 || t >1000)
        cin >> t;
    for (int i=0 ; i<t ; i++){
        cin >> n;
        while (n<1 || t >50)
        cin >> n;
        int arr[n];
        s=0;
        for (int j=0;j<n;j++){
            cin >> arr[j];
            while (arr[j]<1 || arr[j] >100)
                cin >> arr[j];
                if (j % 2 == 1) s=s-arr[j];
            else s=s+arr[j];
        }

        cout << s<<endl;

    }

    }

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

1. **Input Validation for `t` and `n`**:
   - The code incorrectly checks `t > 50` instead of `n > 50` in the inner loop.
   - The input validation for `arr[j]` is not necessary since the problem constraints already specify the valid range for `a_i`.

2. **Edge Cases**:
   - Single element sequence (`n = 1`).
   - Sequence with all elements being the same.
   - Sequence with alternating large and small values.
   - Maximum number of test cases (`t = 1000`).

3. **Test Case Generation**:
   - **Test Case 1**: Single element sequence to check if it handles the simplest case correctly.
   - **Test Case 2**: Sequence with all elements being the same to check if it correctly alternates signs.
   - **Test Case 3**: Sequence with alternating large and small values to check if it correctly handles the alternating sum.
   - **Test Case 4**: Maximum number of test cases with varying lengths to check if it handles the upper limit of `t`.

Here is the test input that covers these scenarios:

```plaintext
4
1
100
3
100 100 100
5
100 1 100 1 100
2
50 50
```

This test input will help expose the bug in the code by:
- Checking the simplest case with a single element.
- Verifying the alternating sum with a sequence of identical elements.
- Testing the alternating sum with a sequence of alternating large and small values.
- Ensuring the code can handle multiple test cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

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

void solve(){
   int n;
   cin>>n;
   int var = 0;
   int ans = 0;
   for(int i=0;i<n;i++){
    int x;
    cin>>x;
    if(var == 0){
        ans += x;
        var =1;
    }else{
        ans -= x;
        var = 0;
    }
   }
   cout<<ans<<endl;
}
int32_t main(){
    int tests;
    cin>>tests;
    for(int i=0;i<tests;i++){
        if(i+1 == 93){
            cout<< 1 <<endl;
        }
         solve();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
import java.util.Scanner;

public class AlteringSumsOfNumbers {

    public static void main(String[] args) {

        Scanner scan = new Scanner(System.in);

        int t = scan.nextInt();
        int[] respuesta = new int[105];

        for (int i = 0; i < t; i++) {
            int a = scan.nextInt();
            int contar = 0;

            for (int j = 0; j < a; j++) {
                if (j % 2 != 0) {
                    contar = contar - scan.nextInt();
                } else {
                    contar = contar + scan.nextInt();
                }
            }
            respuesta[i] = contar;
        }

        for (int i = 0; i < t; i++) {
            System.out.println(respuesta[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
1
50
2
50 50
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: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
	int t;
	int a[10010], answer[110];
	cin >> t;
	for(int i = 0; i < t; i++){
		int n;
		cin >> n;
		answer[i] = 0;
		for(int j = 1; j <= n; j++){
			cin >> a[j];
			if(j % 2 == 1){
				answer[i] = answer[i] + a[j];
			}
			else{
				answer[i] = answer[i] - a[j];
			}
		}
	}
	for(int i = 0; i < t; i++){
		cout << answer[i] << 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
1
50
2
50 50
4
50 50 50 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
#include<bits/stdc++.h>
#define IOS ios_base::sync_with_stdio(false);

using namespace std;
using ll = long long;
int main(){

	int t,n,sum=0, c=-1;
	cin>>t;
	while(t--){
		cin>>n;
	 for(int i=1; i<=n;i++){
		int x;
		cin>>x;
		if(sum == 0){
			sum=x;
			c=-1;
		}else if(c==1){
	     sum+=x;
		  c=-1;
		}else if(c==-1){
		  sum-=x;
			c=1;
		}
	 }
		cout<<sum<<endl;
		sum=0;
	}
	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
1
50
2
50 50
4
50 1 50 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
#include <iostream>

using namespace std;

int main()
{
    int m;
    int in[100][100]={0};
    cin>>m;
    for(int i=0;i<m;i++)
    {
        int n;
        cin>>n;
        for(int j=0;j<n;j++)
        {

            cin>>in[i][j];
        }
    }
    for(int x=0;x<m;x++)
    {
        int sum=0;
        int diff=0;
        for(int y=0;in[x][y]!=0;y=y+2)
        {
            sum=sum+in[x][y];
        }
        for(int y=1;in[x][y]!=0;y=y+2)
        {
            diff=diff+in[x][y];
        }
        cout<<sum-diff<<endl;

    }

}

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

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

Think step by step.

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

1. **Understanding the Bug**: The buggy code reads the input into a 2D array and then calculates the alternating sum by iterating through the array. The main issue is that it uses the condition `in[x][y] != 0` to determine the end of the sequence, which can lead to incorrect results if the sequence contains zeros or if the sequence length is not even.

2. **Edge Cases**:
   - **Sequences with Zeros**: The code might fail if the sequence contains zeros because it uses `in[x][y] != 0` to determine the end of the sequence.
   - **Odd-length Sequences**: The code might fail if the sequence length is odd because the last element might not be correctly included in the sum.
   - **Single-element Sequences**: The code might fail if the sequence has only one element.
   - **Long Sequences**: The code might fail if the sequence length is at the upper limit (50).

3. **Constructing the Test Input**:
   - **Test Case 1**: A sequence with a zero in the middle.
   - **Test Case 2**: A sequence with an odd length.
   - **Test Case 3**: A sequence with a single element.
   - **Test Case 4**: A sequence with the maximum length (50).

Here is the test input that covers these edge cases:

```plaintext
4
5
1 2 0 4 5
3
1 2 3
1
100
50
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
```

This test input includes:
- A sequence with a zero in the middle.
- A sequence with an odd length.
- A sequence with a single element.
- A sequence with the maximum length (50).

These test cases 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: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cmath>
using namespace std;
#define ll long long
void fastIO(void)
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}
void solve()
{
    int n;
    cin >> n;
    int arr[51] = {};
    int pre[51] = {};
    for (int i = 1; i <= n; i++)
    {
        cin >> arr[i];
    }
    int j = 1;
    for (int i = 1; i <= n; i++)
    {
        pre[i] = arr[j] - arr[j + 1];
        j += 2;
    }
    int sum = 0;
    for (int i = 1; i <= (n / 2) + 1; i++)
    {
        sum += pre[i];
    }
    cout << sum << "\n";
}

int main()
{
    fastIO();
    int t = 1;
    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
4
4
1 2 3 4
1
100
2
100 100
5
3 1 4 1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
#include <iostream>
#include <math.h>

using namespace std;

int main (){
	int a, b;
	int sum1 = 0, sum2 = 0;
	cin >> a;
	int arr1[a], arr2[b];
	for (int i = 0; i < a; i++){
		cin >> b;
		for (int i = 0; i < b; i++)
			cin >> arr2[i];
		for (int i = 0; i < b; i++){
			if (i == 0 || i % 2 == 0){
				sum1 += arr2[i];
			}
			else if (i % 2 == 1){
				sum2 += arr2[i];
			}
		}
		arr1[i] = sum1 - sum2;
		sum1 = sum2 = 0;
	}

	for (int i = 0; i < a; i++){
		cout << arr1[i] << 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
1
100
2
1 1
3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
t = int(input())
t1 = 0
while t1 != t:
    t2 = int(input())
    if t2 == 1:
        a = int(input())
        print(a)
    elif t2 == 2:
        a,b = input().split(" ")
        a = int(a)
        b = int(b)
        print(a-b)
    elif t2 == 3:
        a,b,c = input().split(" ")
        a = int(a)
        b = int(b)
        c = int(c)
        print(a-b+c)
    elif t2 == 4:
        a,b,c,d = input().split(" ")
        a = int(a)
        b = int(b)
        c = int(c)
        d = int(d)
        print(a-b+c-d)
    else:
        a,b,c,d,e = input().split(" ")
        a = int(a)
        b = int(b)
        c = int(c)
        d = int(d)
        e = int(e)
        print(a-b+c-d+e)

    t1 = t1 + 1

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The code provided has a fixed structure for handling up to 5 elements, but it does not handle sequences with more than 5 elements correctly. Additionally, it does not handle sequences with fewer than 5 elements efficiently in a general way.

Here are the steps to create a test input:

1. **Test with a single element**: This should work correctly as it is handled separately.
2. **Test with exactly 5 elements**: This should work correctly as it is handled separately.
3. **Test with more than 5 elements**: This is where the bug will likely be exposed.
4. **Test with sequences of different lengths**: To ensure the code handles all possible sequence lengths.

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

```plaintext
4
1
100
5
3 1 4 1 5
6
1 2 3 4 5 6
4
10 20 30 40
```

Explanation:
- The first test case has a single element, which should output `100`.
- The second test case has exactly 5 elements, which should output `10 - 1 + 4 - 1 + 5 = 17`.
- The third test case has 6 elements, which should output `1 - 2 + 3 - 4 + 5 - 6 = -3`.
- The fourth test case has 4 elements, which should output `10 - 20 + 30 - 40 = -20`.

This test input will help expose the bug in the code for sequences with more than 5 elements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
def ivar():                             # reads line as one integer
    return int(input())

def ivars():                            # reads line as list of integers
    return map(int, input().split())        # (map can only be used once, not stored like list)

def ilist():                            # reads line as a list
    return list(ivars())                    # (good for when you want to iterate over the list)

for _ in range(ivar()):                 # repeats for the number of test cases, vertically

    n = ivar()
    s = list(ivars())
    if len(s) == 3:
        print(s[0] - s[1] + s[2])
    if len(s) == 1:
        print(s[0])
    if len(s) == 2:
        print(s[0] - s[1])
    if len(s) == 4:
        print(s[0] - s[1] + s[2] - s[3])
    if len(s) == 5:
        print(s[0] - s[1] + s[2] - s[3] + s[4])

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The main issue with the buggy code is that it only handles specific lengths of the sequence (3, 1, 2, 4, 5) and does not handle other lengths. Therefore, we should create a test case with a sequence length that is not explicitly handled by the code.

Let's consider a sequence length of 6, which is not covered by the code. We will also include a few other test cases to ensure the code is robust.

Here is the test input:

```plaintext
5
6
1 2 3 4 5 6
4
10 20 30 40
3
100 200 300
1
50
2
1 1
```

Explanation:
1. The first test case has a sequence length of 6, which is not handled by the code.
2. The second test case has a sequence length of 4, which is handled by the code.
3. The third test case has a sequence length of 3, which is handled by the code.
4. The fourth test case has a sequence length of 1, which is handled by the code.
5. The fifth test case has a sequence length of 2, which is handled by the code.

This input will help expose the bug in the code by showing that it fails to handle sequences of length 6 correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
import java.util.Scanner;

/**
 * A doc.
 */
public class Main {

  /**
   * A doc.
   */
  public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    int t = scan.nextInt();
    while (t-- > 0) {
      int n = scan.nextInt();
      int[] a = new int[5];
      for (int i = 0; i < n; ++i) {
        a[i] = scan.nextInt();
      }
      int res = 0;
      for (int i=0;i<n;i+=2)
        res+=a[i];
      for(int i=1;i<n;i+=2){
        res-=a[i];
      }
      System.out.println(res);
    }
  }

}

**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
1
100
2
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating Sum of Numbers
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a sequence of integers. Output the alternating sum of this sequence. In other words, output $$$a_1 - a_2 + a_3 - a_4 + a_5 - \dots$$$. That is, the signs of plus and minus alternate, starting with a plus.

Input: The first line of the test contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Then follow $$$t$$$ test cases.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the sequence. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

Output: Output $$$t$$$ lines. For each test case, output the required alternating sum of the numbers.

Examples:
input:
4
4
1 2 3 17
1
100
2
100 100
5
3 1 4 1 5
output:
-15
100
0
10

Note:

Buggy Code:
import java.util.Scanner;

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

        int test_number = inpt.nextInt();
        //int tresult;

        for (int i = 0; i < test_number; i++)
        {
            int test_cases = inpt.nextInt();
            int this_array[] = new int[100];
            int tresult = 0;

            for (int j = 0; j < test_cases; j++)
            {
                this_array[j] = inpt.nextInt();
            }
            for (int j = 0; j < test_cases; j++)
            {
                tresult = this_array[0] - this_array[1] + this_array[2] - this_array[3] + this_array[4] - this_array[5] + this_array[6] - this_array[7] + this_array[8] - this_array[9] + this_array[10] - this_array[11] + this_array[12] - this_array[13] + this_array[14] - this_array[15];
            }
            System.out.println(tresult);
        }

    }
}

**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
4
1
100
2
100 100
3
1 2 3
4
1 2 3 4
```