**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Question Marks
time_limit_ms: 1000
memory_limit_mb: 256
Description: Tim is doing a test consisting of $4n$ questions; each question has $4$ options: 'A', 'B', 'C', and 'D'. For each option, there are exactly $n$ correct answers corresponding to that option — meaning there are $n$ questions with the answer 'A', $n$ questions with the answer 'B', $n$ questions with the answer 'C', and $n$ questions with the answer 'D'.

For each question, Tim wrote his answer on the answer sheet. If he could not figure out the answer, he would leave a question mark '?' for that question.

You are given his answer sheet of $4n$ characters. What is the maximum number of correct answers Tim can get?

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

The first line of each test case contains an integer $n$ ($1 \le n \le 100$).

The second line of each test case contains a string $s$ of $4n$ characters ($s_i \in \\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{?}\\}$) — Tim's answers for the questions.

Output: For each test case, print a single integer — the maximum score that Tim can achieve.

Examples:
input:
6
1
ABCD
2
AAAAAAAA
2
AAAABBBB
2
????????
3
ABCABCABCABC
5
ACADC??ACAC?DCAABC?C
output:
4
2
4
0
9
13

Note: In the first test case, there is exactly one question with each answer 'A', 'B', 'C', and 'D'; so it's possible that Tim gets all his answers correct.

In the second test case, there are only two correct answers 'A' which makes him get exactly $2$ points in any case.

In the third test case, Tim can get at most $2$ correct answers with option 'A' and $2$ correct answers with option 'B'. For example, he would get $4$ points if the answers were 'AACCBBDD'.

In the fourth test case, he refuses to answer any question at all, which makes him get $0$ points.

Buggy Code:
import java.util.*;

public class Questions {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int totalRuns = scanner.nextInt();
        scanner.nextLine();

        for(int i = 0; i < totalRuns; i++) {
            int n = scanner.nextInt();
            scanner.nextLine();
            String answers = scanner.nextLine();
            int A = 0;
            int B = 0;
            int C = 0;
            int D = 0;
            int eazye = 0;
            int incorrect = 0;

            for(int q = 0; q < 4 * n; q++) {
                if(answers.charAt(q) == 'A') {
                    A++;
                }
                if(answers.charAt(q) == 'B') {
                    B++;
                }
                if(answers.charAt(q) == 'C') {
                    C++;
                }
                if(answers.charAt(q) == 'D') {
                    D++;
                }
                if(answers.charAt(q) == '?') {
                    eazye++;
                }
            }

            if(A > n) {
                while(A > n) {
                    incorrect++;
                    A--;
                }
            }
            if(B > n) {
                while(B > n) {
                    incorrect++;
                    B--;
                }
            }
            if(C > n) {
                while(C > n) {
                    incorrect++;
                    C--;
                }
            }
            if(D > n) {
                while(D > n) {
                    incorrect++;
                    A--;
                }
            }
            System.out.println(4 * n - eazye - incorrect);
        }
        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
1
1
A?CD
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Question Marks
time_limit_ms: 1000
memory_limit_mb: 256
Description: Tim is doing a test consisting of $4n$ questions; each question has $4$ options: 'A', 'B', 'C', and 'D'. For each option, there are exactly $n$ correct answers corresponding to that option — meaning there are $n$ questions with the answer 'A', $n$ questions with the answer 'B', $n$ questions with the answer 'C', and $n$ questions with the answer 'D'.

For each question, Tim wrote his answer on the answer sheet. If he could not figure out the answer, he would leave a question mark '?' for that question.

You are given his answer sheet of $4n$ characters. What is the maximum number of correct answers Tim can get?

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

The first line of each test case contains an integer $n$ ($1 \le n \le 100$).

The second line of each test case contains a string $s$ of $4n$ characters ($s_i \in \\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{?}\\}$) — Tim's answers for the questions.

Output: For each test case, print a single integer — the maximum score that Tim can achieve.

Examples:
input:
6
1
ABCD
2
AAAAAAAA
2
AAAABBBB
2
????????
3
ABCABCABCABC
5
ACADC??ACAC?DCAABC?C
output:
4
2
4
0
9
13

Note: In the first test case, there is exactly one question with each answer 'A', 'B', 'C', and 'D'; so it's possible that Tim gets all his answers correct.

In the second test case, there are only two correct answers 'A' which makes him get exactly $2$ points in any case.

In the third test case, Tim can get at most $2$ correct answers with option 'A' and $2$ correct answers with option 'B'. For example, he would get $4$ points if the answers were 'AACCBBDD'.

In the fourth test case, he refuses to answer any question at all, which makes him get $0$ points.

Buggy Code:
from random import *
for _ in range(int(input())):
    n=int(input())
    s=input()
    s=s.replace('?','jamblooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooatautautau')
    fudge_factor_1 = 0
    fudge_factor_2 = 0
    for i in range(8888):
        if random()<0.0000001:
            fudge_factor_1 += 1
        if random()<0.0000001:
            fudge_factor_2 += 1
    print(min(s.count('A')+s.count('?'),n)+min(s.count('B')+s.count('?'),n)+min(s.count('C')+s.count('?'),n)+min(s.count('D')+s.count('?'),n)-fudge_factor_1 + fudge_factor_2)

**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 following aspects:

1. **Edge Cases**: Inputs that are at the boundaries of the constraints.
2. **Randomness and Fudge Factors**: Inputs that can trigger the random fudge factors in the code.
3. **Question Marks**: Inputs with a high number of question marks to test the replacement logic.

The buggy code has a few key issues:
- It replaces question marks with a very long string, which can cause issues with the string length.
- It uses random fudge factors, which can introduce unexpected results.

### Test Input

Let's create a test input that:
- Has a small number of test cases to keep the input small.
- Includes a case with a high number of question marks.
- Includes a case where the number of question marks is just enough to make the fudge factors significant.

```plaintext
2
1
????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????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**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Question Marks
time_limit_ms: 1000
memory_limit_mb: 256
Description: Tim is doing a test consisting of $4n$ questions; each question has $4$ options: 'A', 'B', 'C', and 'D'. For each option, there are exactly $n$ correct answers corresponding to that option — meaning there are $n$ questions with the answer 'A', $n$ questions with the answer 'B', $n$ questions with the answer 'C', and $n$ questions with the answer 'D'.

For each question, Tim wrote his answer on the answer sheet. If he could not figure out the answer, he would leave a question mark '?' for that question.

You are given his answer sheet of $4n$ characters. What is the maximum number of correct answers Tim can get?

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

The first line of each test case contains an integer $n$ ($1 \le n \le 100$).

The second line of each test case contains a string $s$ of $4n$ characters ($s_i \in \\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{?}\\}$) — Tim's answers for the questions.

Output: For each test case, print a single integer — the maximum score that Tim can achieve.

Examples:
input:
6
1
ABCD
2
AAAAAAAA
2
AAAABBBB
2
????????
3
ABCABCABCABC
5
ACADC??ACAC?DCAABC?C
output:
4
2
4
0
9
13

Note: In the first test case, there is exactly one question with each answer 'A', 'B', 'C', and 'D'; so it's possible that Tim gets all his answers correct.

In the second test case, there are only two correct answers 'A' which makes him get exactly $2$ points in any case.

In the third test case, Tim can get at most $2$ correct answers with option 'A' and $2$ correct answers with option 'B'. For example, he would get $4$ points if the answers were 'AACCBBDD'.

In the fourth test case, he refuses to answer any question at all, which makes him get $0$ points.

Buggy Code:
from random import *
for _ in range(int(input())):
    n=int(input())
    s=input()
    s=s.replace('?','jamblooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooatautautau')
    fudge_factor_1 = 0
    fudge_factor_2 = 0
    for i in range(8888):
        if random()<0.0000001:
            fudge_factor_1 += 1
        if random()<0.0000001:
            fudge_factor_2 += 1
    print(min(s.count('A')+s.count('?'),n)+min(s.count('B')+s.count('?'),n)+min(s.count('C')+s.count('?'),n)+min(s.count('D')+s.count('?'),n)-fudge_factor_1 + fudge_factor_2)

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

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

Think step by step.

Response:
```plaintext
1
5
ACADC??ACAC?DCAABC?C
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Question Marks
time_limit_ms: 1000
memory_limit_mb: 256
Description: Tim is doing a test consisting of $4n$ questions; each question has $4$ options: 'A', 'B', 'C', and 'D'. For each option, there are exactly $n$ correct answers corresponding to that option — meaning there are $n$ questions with the answer 'A', $n$ questions with the answer 'B', $n$ questions with the answer 'C', and $n$ questions with the answer 'D'.

For each question, Tim wrote his answer on the answer sheet. If he could not figure out the answer, he would leave a question mark '?' for that question.

You are given his answer sheet of $4n$ characters. What is the maximum number of correct answers Tim can get?

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

The first line of each test case contains an integer $n$ ($1 \le n \le 100$).

The second line of each test case contains a string $s$ of $4n$ characters ($s_i \in \\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{?}\\}$) — Tim's answers for the questions.

Output: For each test case, print a single integer — the maximum score that Tim can achieve.

Examples:
input:
6
1
ABCD
2
AAAAAAAA
2
AAAABBBB
2
????????
3
ABCABCABCABC
5
ACADC??ACAC?DCAABC?C
output:
4
2
4
0
9
13

Note: In the first test case, there is exactly one question with each answer 'A', 'B', 'C', and 'D'; so it's possible that Tim gets all his answers correct.

In the second test case, there are only two correct answers 'A' which makes him get exactly $2$ points in any case.

In the third test case, Tim can get at most $2$ correct answers with option 'A' and $2$ correct answers with option 'B'. For example, he would get $4$ points if the answers were 'AACCBBDD'.

In the fourth test case, he refuses to answer any question at all, which makes him get $0$ points.

Buggy Code:
from random import *
for _ in range(int(input())):
    n=int(input())
    s=input()
    s=s.replace('?','jamblooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooatautautau')
    fudge_factor_1 = 0
    fudge_factor_2 = 0
    for i in range(10000):
        if random()<0.0000001:
            fudge_factor_1 += 1
        if random()<0.0000001:
            fudge_factor_2 += 1
    print(min(s.count('A')+s.count('?'),n)+min(s.count('B')+s.count('?'),n)+min(s.count('C')+s.count('?'),n)+min(s.count('D')+s.count('?'),n)-fudge_factor_1 + fudge_factor_2)

**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
10
AAAAAAAAAAAAAA??
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Question Marks
time_limit_ms: 1000
memory_limit_mb: 256
Description: Tim is doing a test consisting of $4n$ questions; each question has $4$ options: 'A', 'B', 'C', and 'D'. For each option, there are exactly $n$ correct answers corresponding to that option — meaning there are $n$ questions with the answer 'A', $n$ questions with the answer 'B', $n$ questions with the answer 'C', and $n$ questions with the answer 'D'.

For each question, Tim wrote his answer on the answer sheet. If he could not figure out the answer, he would leave a question mark '?' for that question.

You are given his answer sheet of $4n$ characters. What is the maximum number of correct answers Tim can get?

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

The first line of each test case contains an integer $n$ ($1 \le n \le 100$).

The second line of each test case contains a string $s$ of $4n$ characters ($s_i \in \\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{?}\\}$) — Tim's answers for the questions.

Output: For each test case, print a single integer — the maximum score that Tim can achieve.

Examples:
input:
6
1
ABCD
2
AAAAAAAA
2
AAAABBBB
2
????????
3
ABCABCABCABC
5
ACADC??ACAC?DCAABC?C
output:
4
2
4
0
9
13

Note: In the first test case, there is exactly one question with each answer 'A', 'B', 'C', and 'D'; so it's possible that Tim gets all his answers correct.

In the second test case, there are only two correct answers 'A' which makes him get exactly $2$ points in any case.

In the third test case, Tim can get at most $2$ correct answers with option 'A' and $2$ correct answers with option 'B'. For example, he would get $4$ points if the answers were 'AACCBBDD'.

In the fourth test case, he refuses to answer any question at all, which makes him get $0$ points.

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

using ll = long long;
using db = double;
using vll = vector<long long>;
using vpll = vector<pair<long long, long long>>;
using vvll = vector<vector<long long>>;
using gll = greater<long long>;
using pll = pair<ll, ll>;
#define rep(i, a, n) for (int i = a; i < n; i++)
#define pb push_back
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

#define fast                     \
    ios::sync_with_stdio(false); \
    cin.tie(0);

ll gcd(ll a, ll b) {
    while (b != 0) {
        ll tmp = a % b;
        a = b;
        b = tmp;
    }
    return a;
}
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }

void print(ll* a, int n) {
    for (int i = 0; i < n; i++) {
        cout << a[i] << " ";
    }
    cout << endl;
}

void print(vll& a) {
    for (int i = 0; i < a.size(); i++) {
        cout << a[i] << " ";
    }
    cout << endl;
}

void solve() {
    int n;
    string s;
    cin >> n>>s;
    map<int,char> mp;
    for(int i=0; i<4*n;i++){
        mp[s[i]]++;
    }
    int a=mp['A'],b=mp['B'],c=mp['C'],d=mp['D'],left=mp['?'];
    int res=0;
    res+=min(a,n);
    res+=min(b,n);
    res+=min(c,n);
    res+=min(d,n);
    cout<<res<<endl;
}

int main() {
    //(^w^)//
    fast int sura = 1;
    cin >> sura;
    while (sura--) {
        solve();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the code might not handle the distribution of question marks correctly. The code currently counts the number of each answer ('A', 'B', 'C', 'D') and question marks ('?'), and then sums the minimum of each answer count and `n`. However, it does not account for the possibility of using question marks to maximize the score.

### Step-by-Step Analysis:
1. **Correct Distribution of Answers**:
   - The code should ensure that the number of correct answers for each option ('A', 'B', 'C', 'D') is exactly `n`.

2. **Handling Question Marks**:
   - The code should use question marks to fill in the gaps where the number of correct answers is less than `n`.

3. **Test Case Construction**:
   - We need a test case where the number of correct answers for some options is less than `n`, and there are enough question marks to fill these gaps.

### Example Test Case:
Let's construct a test case where:
- `n = 2`
- The string `s` contains fewer correct answers for some options and some question marks.

```plaintext
1
2
AABBC?
```

### Explanation:
- `n = 2`: There are 2 questions for each correct answer ('A', 'B', 'C', 'D').
- `s = "AABBC?"`:
  - 'A' appears 2 times.
  - 'B' appears 2 times.
  - 'C' appears 1 time.
  - 'D' appears 0 times.
  - There is 1 question mark.

The correct answer should be:
- 2 correct answers for 'A'.
- 2 correct answers for 'B'.
- 1 correct answer for 'C' (since there is 1 'C' and 1 question mark that can be used for 'C').
- 1 correct answer for 'D' (since there is 1 question mark left that can be used for 'D').

Thus, the maximum score Tim can achieve is `2 + 2 + 1 + 1 = 6`.

### Output:
```plaintext
1
2
AABBC?
```

This test case should expose the bug in the code, as the current implementation does not account for the distribution of question marks to maximize the score.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Question Marks
time_limit_ms: 1000
memory_limit_mb: 256
Description: Tim is doing a test consisting of $4n$ questions; each question has $4$ options: 'A', 'B', 'C', and 'D'. For each option, there are exactly $n$ correct answers corresponding to that option — meaning there are $n$ questions with the answer 'A', $n$ questions with the answer 'B', $n$ questions with the answer 'C', and $n$ questions with the answer 'D'.

For each question, Tim wrote his answer on the answer sheet. If he could not figure out the answer, he would leave a question mark '?' for that question.

You are given his answer sheet of $4n$ characters. What is the maximum number of correct answers Tim can get?

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

The first line of each test case contains an integer $n$ ($1 \le n \le 100$).

The second line of each test case contains a string $s$ of $4n$ characters ($s_i \in \\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{?}\\}$) — Tim's answers for the questions.

Output: For each test case, print a single integer — the maximum score that Tim can achieve.

Examples:
input:
6
1
ABCD
2
AAAAAAAA
2
AAAABBBB
2
????????
3
ABCABCABCABC
5
ACADC??ACAC?DCAABC?C
output:
4
2
4
0
9
13

Note: In the first test case, there is exactly one question with each answer 'A', 'B', 'C', and 'D'; so it's possible that Tim gets all his answers correct.

In the second test case, there are only two correct answers 'A' which makes him get exactly $2$ points in any case.

In the third test case, Tim can get at most $2$ correct answers with option 'A' and $2$ correct answers with option 'B'. For example, he would get $4$ points if the answers were 'AACCBBDD'.

In the fourth test case, he refuses to answer any question at all, which makes him get $0$ points.

Buggy Code:
//
//  main.cpp
//  zy_123
//
//  Created by 燃烧星空 on 2024/1/20.
//
//
//#include<bits/stdc++.h>
#include<iostream>
#include<map>
#include<unordered_map>
#include<set>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<string>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<unordered_set>
#include<math.h>
#include<stack>
#include<stdint.h>
#include<bitset>
#include<fstream>
#include<random>
#include<time.h>
#define IOS cin.tie(0) -> sync_with_stdio(false), cout.tie(0);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=1e6+5;
const int M=1e6+5;
const int inf=1e8;
const ll INF=0x3f3f3f3f;
#define lc p<<1
#define rc p<<1|1
#define lb(x) (x&(-x))
#define endl "\n"
double pai = acos(-1);
const ll mod=1e9+7;
//__int128 read_128()//c++20(64)可用
//{
//    __int128 x=0,f=1;char ch = getchar();
//    while(ch < '0' || ch > '9'){if(ch == '-') f=-1;ch=getchar();}
//    while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch=getchar();}
//    return x*f;
//}
//void print_128(__int128 x)
//{
//    if(x<0){putchar('-');x=-x;}
//    if(x>9)print_128(x/10);
//    putchar(x % 10 + '0');
//}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    for ( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
    return x * f;
}
int idx=1,nxt[M*2],dian[M*2],h[N];
ll quan[M*2];
//ll rong[M*2];
int cur[N];
void add(int a,int b,ll c)
{
    dian[++idx]=b;
    quan[idx]=c;
    //rong[idx]=c;
    nxt[idx]=h[a];
    h[a]=idx;
}
ll ksm(ll a,ll b){ll sum=1;while(b){if(b&1){sum*=a;}a*=a;b>>=1;}return sum;}
ll ksm_mod(ll a,ll b,ll p){ll sum=1;while(b){if(b&1){sum=sum*a%p;}a=a*a%p;b>>=1;}return sum%p;}
ll gsc(ll a,ll b,ll p){ll sum=0;while(b){if(b&1){sum=(sum+a)%p;}a=a*2%p;b>>=1;}return sum%p;}
ll quick_mul(ll x,ll y,ll p){ll z=(long double)x/p*y;ll res=(ull)x*y-(ull)z*p;return (res+p)%p;}
//mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());

int n,m;
ll a[N];
char s[5];
void solve()
{
    cin >> n;
    string ss;
    cin >> ss;
    for(int i=0;i<4;i++)
    {
        s[i]=0;
    }
    for(int i=0;i<n*4;i++)
    {
        if(ss[i]=='?')
        {
            continue;
        }
        int x=ss[i]-'A';
        s[x]++;
    }
    ll ans=0;
    for(int i=0;i<4;i++)
    {
        ans+=min((int)s[i],n);
    }
    cout << ans << "\n";
}
int main()
{
    IOS;
    int t=1;
    cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: Test cases with the minimum and maximum values of `n` (1 and 100).
2. **Special Characters**: Test cases with a mix of 'A', 'B', 'C', 'D', and '?'.
3. **Imbalance**: Test cases where the distribution of 'A', 'B', 'C', and 'D' is not equal, but still within the constraints.

The buggy code does not handle the case where the number of each character (A, B, C, D) exceeds the number of correct answers (n). It simply sums up the minimum of the counts of each character and `n`, which can lead to incorrect results if the counts are not balanced.

Let's create a test input that maximizes the failure probability:

```plaintext
4
1
ABCD
2
AAABBBCCCD
3
AAAAABBBBCCCCDDDD
100
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
```

### Explanation:
1. **Test Case 1**: `n = 1`, `s = "ABCD"`
   - This is a simple case where the input is balanced and the expected output is 4.
2. **Test Case 2**: `n = 2`, `s = "AAABBBCCCD"`
   - This case has an imbalance with more 'A's and 'B's than 'C's and 'D's. The expected output is 8.
3. **Test Case 3**: `n = 3`, `s = "AAAAABBBBCCCCDDDD"`
   - This case has an imbalance with more 'A's and 'B's than 'C's and 'D's. The expected output is 12.
4. **Test Case 4**: `n = 100`, `s = "A" * 400`
   - This case has a large number of 'A's. The expected output is 100, but the buggy code might fail due to the imbalance.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Question Marks
time_limit_ms: 1000
memory_limit_mb: 256
Description: Tim is doing a test consisting of $4n$ questions; each question has $4$ options: 'A', 'B', 'C', and 'D'. For each option, there are exactly $n$ correct answers corresponding to that option — meaning there are $n$ questions with the answer 'A', $n$ questions with the answer 'B', $n$ questions with the answer 'C', and $n$ questions with the answer 'D'.

For each question, Tim wrote his answer on the answer sheet. If he could not figure out the answer, he would leave a question mark '?' for that question.

You are given his answer sheet of $4n$ characters. What is the maximum number of correct answers Tim can get?

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

The first line of each test case contains an integer $n$ ($1 \le n \le 100$).

The second line of each test case contains a string $s$ of $4n$ characters ($s_i \in \\{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, \texttt{?}\\}$) — Tim's answers for the questions.

Output: For each test case, print a single integer — the maximum score that Tim can achieve.

Examples:
input:
6
1
ABCD
2
AAAAAAAA
2
AAAABBBB
2
????????
3
ABCABCABCABC
5
ACADC??ACAC?DCAABC?C
output:
4
2
4
0
9
13

Note: In the first test case, there is exactly one question with each answer 'A', 'B', 'C', and 'D'; so it's possible that Tim gets all his answers correct.

In the second test case, there are only two correct answers 'A' which makes him get exactly $2$ points in any case.

In the third test case, Tim can get at most $2$ correct answers with option 'A' and $2$ correct answers with option 'B'. For example, he would get $4$ points if the answers were 'AACCBBDD'.

In the fourth test case, he refuses to answer any question at all, which makes him get $0$ points.

Buggy Code:
import java.io.DataInputStream;
import java.io.IOException;
import java.io.InputStream;

public class Main {
	Main(Parser in) {
		int n = in.readInt();
		char[] a = in.readCharArray();

		int nA = 0, nB = 0, nC = 0, nD = 0, nQ = 0;

		for (char c : a) {
			if (c == 'A')
				nA++;
			if (c == 'B')
				nB++;
			if (c == 'C')
				nC++;
			if (c == 'D')
				nD++;
			if (c == '?')
				nQ++;
		}

		int w =
				Math.max(0, nA - n) +
						Math.max(0, nB - n) +
						Math.max(0, nC - n) +
						Math.max(0, nD - n);

		System.out.println(4 * n - nQ - w);
	}

	public static void main(String[] args) {
		Parser in = new Parser();
		int nTests = in.readInt();
		for (int test = 0; test < nTests; test++) {
			new Main(in);
		}
	}

	private static class Parser {
		final private int BUFFER_SIZE = 1 << 16;
		private final DataInputStream dis;
		private final byte[] buffer;
		private int bufferPointer, bytesRead;

		public Parser(InputStream is) {
			dis = new DataInputStream(is);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
			fillBuffer();
		}

		public Parser() {
			this(System.in);
		}

		private byte nextByte;

		private void fillBuffer() {
			try {
				bytesRead = dis.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
			if (bytesRead == -1) {
				buffer[0] = -1;
			}
			nextByte = buffer[bufferPointer];
		}

		private void advance() {
			bufferPointer++;
			if (bufferPointer == bytesRead) {
				fillBuffer();
			}
		}

		private byte next() {
			return nextByte;
		}

		private void consume() {
			advance();
			nextByte = buffer[bufferPointer];
		}

		private byte read() {
			advance();
			byte b = nextByte;
			nextByte = buffer[bufferPointer];
			return b;
		}

		private boolean isWhitespace(byte b) {
			return b <= 32;
		}

		private boolean isDigit(byte b) {
			return b >= '0' && b <= '9';
		}

		private void skipWhitespace() {
			while (isWhitespace(next())) {
				consume();
			}
		}

		public char[] readCharArray() {
			skipWhitespace();

			ArrayBytes buf = new ArrayBytes(64);
			while (!isWhitespace(next())) {
				buf.add(read());
			}
			return buf.asCharArrayASCII();
		}

		public int readInt() {
			return Math.toIntExact(readLong());
		}

		public long readLong() {
			skipWhitespace();

			boolean negative = (next() == '-');
			if (negative) {
				consume();
			}

			long m = 0;
			while (isDigit(next())) {
				m = m * 10 + read() - '0';
			}

			return negative ? -m : m;
		}
	}

	private static class ArrayBytes {
		int n;
		byte[] values;
		int capacity;
		boolean backed;

		public ArrayBytes(int initialCapacity) {
			if (initialCapacity < 1) {
				throw new IllegalArgumentException("Initial capacity must be at least 1.");
			}
			capacity = initialCapacity;
			values = new byte[capacity];
			n = 0;
			backed = false;
		}

		public ArrayBytes() {
			this(8);
		}

		void ensureCapacityGreaterThan(int capacityWanted) {
			if (capacity <= capacityWanted) {
				if (backed) {
					throw new UnsupportedOperationException("Can not extend backed ArrayBytes.");
				}
				capacity = Integer.highestOneBit(capacityWanted) << 1;
				byte[] valuesNew = new byte[capacity];
				System.arraycopy(values, 0, valuesNew, 0, n);
				values = valuesNew;
			}
		}

		public void add(byte value) {
			ensureCapacityGreaterThan(n);
			values[n++] = value;
		}

		public char[] asCharArrayASCII() {
			char[] chars = new char[n];
			for (int i = 0; i < n; i++) {
				chars[i] = (char) values[i];
			}
			return chars;
		}
	}
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the algorithm might fail to correctly calculate the maximum number of correct answers. The bug in the code is that it incorrectly calculates the number of extra answers for each option ('A', 'B', 'C', 'D') and then subtracts these from the total possible correct answers. This approach fails when the number of '?' characters can be used to balance out the excess answers.

Let's consider a test case where the number of '?' characters is crucial to the correct solution. Specifically, we can create a test case where the number of each option ('A', 'B', 'C', 'D') is just one more than the expected number of correct answers for that option, and the rest of the answers are '?'.

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

```plaintext
1
3
AAABBBCCCDDD????
```

### Explanation:
- **n = 3**: There are 3 correct answers for each option ('A', 'B', 'C', 'D').
- **String "AAABBBCCCDDD????"**:
  - There are 3 'A's, 3 'B's, 3 'C's, and 3 'D's.
  - There are 7 '?' characters.

### Expected Output:
- The maximum number of correct answers Tim can get is 12.
- The 7 '?' characters can be used to fill in the missing correct answers for 'A', 'B', 'C', and 'D'.

### Why This Test Case Exposes the Bug:
- The code incorrectly calculates the number of extra answers for each option and subtracts them from the total possible correct answers.
- The correct approach should be to use the '?' characters to balance out the excess answers and maximize the number of correct answers.

This test case should demonstrate the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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<iostream>
#include<string>
using namespace std;
int main()
{
	char note;
	int numc,i,num,flag=0,flag1=0;
	string a;
	cin>>numc;
	while(numc--)
	{
		cin>>num;
		cin>>a;
		for(i=0;i<a.size();i++)
		{
			if(a[i]!='.'&&flag==0)
			{
				flag=1;
				note=a[i];
			}
			if(flag)
			{
				if(a[i]!=note&&i!=a.size()-1&&a[i]!='.')
				{
					flag1=1;
					break;
				}
			}
		}
		if(flag1) cout<<"NO"<<endl;
		else cout<<"YES"<<endl;
		flag=0;flag1=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
4
spsp
5
psps.
6
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:
#include<iostream>
using namespace std;
int main() {
	int t = 0;
	cin >> t;
	while (t--) {
		int n = 0;
		string s;
		cin >> n >> s;
		int cnt = 0, det = 0;
		int xs = 0, ys = 0;
		int x = 0, y = 0;
		int dx = 0, dy = 0;
		for (int i = 0; i < s.length(); i++) {
			cnt++;
			if (s[i] == 'p') {
				if (!xs)xs = cnt;
				x = cnt;
				dx = 1;
			}
			else if (s[i] == '.');
			else;
		}
		cnt = 0;
		for (int i = s.length() - 1; i >= 0; i--) {
			cnt++;
			if (s[i] == 's') {
				if (!ys)ys = cnt;
				y = cnt;
				dy = 1;
			}
			else if (s[i] == '.');
			else;
		}
		if (dx && dy) {
			int len = s.length();
			y = len - y + 1;
			ys = len - ys + 1;
			if ((y == 1 && x == len))det = 0;
			else det = 1;
			if (!det)cout << "YES" << endl;
			else cout << "NO" << endl;
		}
		else {
			cout << "YES" << endl;
		}
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
8
pspspsps
```


**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;
using namespace chrono;

void solver() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int input; cin >> input;
    while(input--) {
        int n; cin >> n;
        string str; cin >> str;
        bool s = false, possi = true;
        vector<int> vec(str.size());
        int lastP = 0;

        for(int i = str.size() - 1; i >= 0; i--) {
            if(str[i] == 'p' && s) {
                possi = false;
                break;
            }
            if(str[i] == 'p') lastP = i + 1;
            if(str[i] == 's') {
                int temp = i + 1 + str.size() - i - 1;
                if(lastP && lastP - temp < 0) {
                    possi = false;
                    break;
                }
            }
            if(str[i] == 's') s = true;
        }

        if(possi) cout << "YES\n";
        else cout << "NO\n";
    }
}

int main(){

    solver();

}

**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 conditions and constraints of the problem:

1. **Permutation Requirements**:
   - If `s_i` is `p`, the prefix `[p_1, p_2, ..., p_i]` must be a permutation.
   - If `s_i` is `s`, the suffix `[p_i, p_i+1, ..., p_n]` must be a permutation.
   - If `s_i` is `.`, there is no additional restriction.

2. **Buggy Code Analysis**:
   - The code seems to check if a `p` appears after an `s` and sets `possi` to `false` if it does.
   - It also checks the difference between the positions of `p` and `s` and sets `possi` to `false` if the difference is invalid.
   - However, the logic for checking the difference between positions might be flawed.

3. **Edge Cases**:
   - **Empty String**: Not applicable since the minimum length is 1.
   - **All `p` or All `s`**: These should be straightforward and should work correctly.
   - **Mixed `p` and `s`**: This is where the bug might be more visible.
   - **Longer Strings with Multiple `p` and `s`**: This can help identify issues with the position checks.

4. **Challenging Test Case**:
   - A string with a mix of `p` and `s` where the positions are such that the permutation requirements are violated but the code might not catch it.

Given the constraints, a small-scale test input that could expose the bug is:

```plaintext
4
4
spsp
5
pspsp
6
pspsps
7
spspsps
```

This test input includes strings with mixed `p` and `s` that should be challenging for the buggy code to handle correctly. The expected output for these test cases is:

- For `spsp`: `NO` (The prefix and suffix conditions cannot be satisfied simultaneously)
- For `pspsp`: `NO` (Similar to the above)
- For `pspsps`: `NO` (Similar to the above)
- For `spspsps`: `NO` (Similar to the above)

This 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: 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:
t = int(input())
for _ in range(t):
    n = int(input())
    s = input()

    if len(s) == 0:
      print("YES")
      continue

    for i in range(len(s)):
      if i == len(s) - 1:
        print("YES")
      if s[i] == 'p':
        if 's' not in s[i+1:]:
          print("YES")
        else:
          print("NO")
        break
      elif s[i] == 's':
        index = s.find('p', i+1)
        if index == len(s) - 1 or index == -1:
          print("YES")
          break
        if s.count('s') > 1:
          print("NO")
          break
        if s[0] != 's':
          print("NO")
          break
        print("YES")
        break

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

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

Think step by step.

Response:
```plaintext
5
4
psps
6
pssps.
5
pss..
8
psspsps.
10
pspssps..
```

5
4
psps
6
pssps.
5
pss..
8
psspsps.
10
pspssps..


**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:
#بِسْمِ اللهِ الرَّحْمٰنِ الرَّحِيْمِ
#"In the name of Allah,the Most Gracious,the Most Merciful".
for _ in range(int(input())):
  n=int(input())
  s=input()
  if n==1:print('YES')
  elif len(set(s))==1:print('YES')
  elif 's' in s and s[0]=='s':print('YES')
  else:print('NO')

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

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

Think step by step.

Response:
```plaintext
3
4
s.sp
6
pss..s
8
pspspsps
```


**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:
t = int(input())
for _ in range(t):
    n = int(input())
    s = input()
    s_cnt, p_cnt = 0, 0
    for ele in s:
        if ele == 's':
            s_cnt += 1
        elif ele == 'p':
            p_cnt += 1
    if s_cnt == 0 or p_cnt == 0:
        print("YES")
    else:
        s_list = []
        p_list = []
        for i in range(n):
            if s[i] == 's':
                s_list.append(i)
            elif s[i] == 'p':
                p_list.append(i)
        flag = False
        for i in p_list:
            for j in s_list:
                if i < j:
                    flag = True
                    break
            if flag:
                break
        if flag:
            print('NO')
        else:
            res = [-1] * n
            x = 1
            for i in range(s_list[0], p_list[0] + 1):
                res[i] = x
                x += 1
            transition = 0
            for i in range(n - 1):
                if res[i] == -1 and res[i + 1] != -1:
                    transition += 1
                elif res[i] != -1 and res[i + 1] == -1:
                    transition += 1
            if transition >= 2:
                print('NO')
            else:
                print('YES')

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

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

Think step by step.

Response:
```plaintext
4
4
s.p.
5
ps..s
6
p...s.
7
p...s.s
```

### Explanation:
1. **Test Case 1: `4 s.p.`**
   - This string has a single 's' at the beginning and a single 'p' in the middle.
   - The correct permutation for this case would be `[3, 4, 1, 2]`, but the buggy code might fail to handle the transition between 's' and 'p'.

2. **Test Case 2: `5 ps..s`**
   - This string has a 'p' at the beginning and an 's' at the end, with dots in between.
   - The correct permutation for this case would be `[1, 2, 3, 4, 5]`, but the buggy code might incorrectly handle the transition between 'p' and 's'.

3. **Test Case 3: `6 p...s.`**
   - This string has a 'p' at the beginning and an 's' near the end, with dots in between.
   - The correct permutation for this case would be `[1, 2, 3, 4, 5, 6]`, but the buggy code might fail to handle the transition between 'p' and 's'.

4. **Test Case 4: `7 p...s.s`**
   - This string has a 'p' at the beginning and two 's' characters, with dots in between.
   - The correct permutation for this case would be `[1, 2, 3, 4, 5, 6, 7]`, but the buggy code might incorrectly handle the multiple 's' characters and their transitions.

These test cases are designed to expose the bug in the faulty solution by focusing on edge cases where the transitions between 'p' and 's' are critical.


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

public class Main {
    static FastIO sc = new FastIO();

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

    static void solve() {
        int n=sc.nextInt();
        String s=sc.next();
        int c1=0,c2=0,i=0,cnt1=0,cnt2=0;
        for(i=0;i<n;i++)
        {
            if(s.charAt(i)=='s')
            c1++;
            else if(s.charAt(i)=='p')
            c2++;
        }
        if(c1==0 || c2==0)
        {
            System.out.println("YES");
            return;
        }
        for(i=0;i<n;i++)
        {
            if(s.charAt(i)!='.')
            break;
        }
        cnt1=i;
        for(i=n-1;i>=0;i--)
        {
            if(s.charAt(i)!='.')
            break;
        }
        cnt2=n-i-1;
        if(cnt1>0 && cnt2>0)
        {
            System.out.println("NO");
            return;
        }
        for(i=0;i<n;i++)
        {
            if(c1==0)
            break;
             if(s.charAt(i)=='p')
             {
                System.out.println("NO");
                return;
             }
             if(s.charAt(i)=='s')
             c1--;

        }
        System.out.println("YES");
        return;
    }

    static class Pair {
        long x;
        int y;

        Pair(long x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    static class FastIO extends PrintWriter {
        private BufferedReader br;
        private StringTokenizer st;

        public FastIO() {
            this(System.in, System.out);
        }

        public FastIO(InputStream in, OutputStream out) {
            super(out);
            br = new BufferedReader(new InputStreamReader(in));
        }

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

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

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

**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
.psp
```


**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.*;
public class aa{
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            String s=sc.next();
            int p=0;
            int ss=0;
            boolean flag =true;
            char prev='z';
            for(int i=0;i<s.length();i++){
                if(s.charAt(i)=='p'){
                    prev='p';
                    p++;
                }
                else if(s.charAt(i)=='s'){
                    ss++;
                }
                if(prev=='p'){
                    if(s.charAt(i)=='s'){
                        flag=false;
                        break;
                    }
                }
            }
            if((p==0 || ss==0) && flag){
                System.out.println("YES");
                continue;
            }
            if(flag){
                if(s.charAt(0)=='s' || s.charAt(n-1)=='p') System.out.println("YES");
                else{
                    System.out.println("NO");
                }
        }
        else{
            System.out.println("NO");
        }
    }
}
}

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

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

Think step by step.

Response:
```plaintext
3
5
pssp.
6
pss..s
4
spsp
```


**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;

typedef vector<int> vecy;

int main(){
  int t;
  cin>>t;
  while(t--){
 int n;
 cin>>n;
  string s;
  cin>>s;

  int counts=0,countp=0;
  int countps=0;
  int ans=0;

  for(int i=0;i<n;i++){

    if(countp>0 && s[i]=='s' && counts==0){
       ans++ ;
        break;}
        if(counts>0 && s[i]=='p' && countp==0  && i!=n-1){
       ans++ ;
        break;}
    if(s[i]=='s'){ counts++;}
    if(s[i]=='p') {countp++;}
if(i!=0 && i!=n-1 && i!=1 && s[i]=='p' && s[i-1]=='s'){
  ans++ ;
  break;
}
  }
  if(counts==1 && s[0]=='s'){
    cout<<"YES"<<endl;
    continue;

  }
   if(countp==1 && s[n-1]=='p'){
    cout<<"YES"<<endl;
    continue;

  }

if(ans>0)
{cout<<"NO"<<endl;}else{
  cout<<"YES"<<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
4
spsp
6
pssp..
8
psspss..
20
psspsssspsspsssspss..
```

4
4
spsp
6
pssp..
8
psspss..
20
psspsssspsspsssspss..


**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 endl "\n"
#define vi vector<int>
const int mod=1e9+7;

void solve(){
	 char ch[550];
	 map<char,int> ma;
	 int n;
	 char num;
	 cin>>n;
	 for(int i=1;i<=n;i++){
	 	cin>>ch[i];
	 	if(i>=2&&i<=n-1){
	 		ma[ch[i]]=1;
	 		num=ch[i];
		 }
	 }
	 if(n==1||n==2){
	 	if(n==1) cout<<"yes\n";
	 	if(n==2) cout<<(((ch[1]=='s'&&ch[n]=='p')||ch[1]==ch[n])?"yes\n":"no\n");
	 	return ;
	 }
	 if(ma['s']&&ma['p']){
	 	cout<<"no\n";
	 	return ;
	 }
	 if(num==ch[1]&&num==ch[n]){
	 	cout<<"yes\n";
	 	return ;
	 }
	 if(ch[1]=='s'&&num!='s'&&ch[n]!='s'){
	 	cout<<"yes\n";
	 	return ;
	 }
	 if(ch[n]=='p'&&num!='p'&&ch[1]!='p'){
	 	cout<<"yes\n";
	 	return ;
	 }
	 cout<<"no\n";
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int t;
	cin>>t;

	while(t--){
		solve();
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
3
pps
3
sps
3
pps
3
psp
```


**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:
t=int(input())
for _ in range(t):
    n=int(input())
    s=list(input())
    listd=[]
    if s==["." for _ in range(n)] or s==["s" for _ in range(n)] or s==["p" for _ in range(n)]:
        print("YES")
    else:
        dic1={}
        s=[""]+s
        for i in range(1,len(s)):
            if s[i]=='s':
                l=n-i+1
                if l not in dic1:
                    dic1[l]=[i,n]
                else:
                    if i!=dic1[l][0]:
                        print("NO")
                        break
                if listd==[]:
                    listd.append([i,n])
                else:
                    if (n<listd[0][0]) or (i>listd[0][1]):
                        print("NO")
                        break
            elif s[i]=='p':
                l=i
                if l not in dic1:
                    dic1[l]=[1,i]
                else:
                    if i!=dic1[l][1]:
                        print("NO")
                        break
                if listd==[]:
                    listd.append([1,i])
                    if (i<listd[0][0]) or (1>listd[0][1]):
                        print("NO")
                        break
        else:
            print("YES")

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

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

Think step by step.

Response:
```plaintext
5
4
spsp
5
pspsp
6
pssp..
7
pspspsp
8
psspspsp
```

5
4
spsp
5
pspsp
6
pssp..
7
pspspsp
8
psspspsp


**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:
t=int(input())
for _ in range(t):
    n=int(input())
    s=input()
    ss=list(s)
    x=s.find("p")
    max=0
    y=s.find("s",1)
    while max<y:
        max=y
        y=s.find("s",y+1)
    if x!=-1 and x<max:
        print("NO")
    else:
        x=s.find("sp")
        if x!=-1 and n>2:
            if x==0 or x==n-2:
                print("YES")
            else:
                print("NO")
        else:
            x=s.find("ps")
            if x!=-1:
                print("NO")
            else:
                print("YES")

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

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

Think step by step.

Response:
```plaintext
3
4
psps
5
p..s.
8
pspspsps
```


**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.*;

public class Demo {
    private static String bestString = "";
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        scanner.nextLine();

        while(t-- > 0) {
            int n = scanner.nextInt();
            scanner.nextLine();
            String s = scanner.nextLine();
            //System.out.println(s);
            int l = 0, r =n - 1;
            int ls = 0, rs = 0;
            int lp = 0, rp = 0;
            boolean ans = true;
            boolean leadOrTrailingZeros = false;

            if(s.charAt(0) == '.' && s.charAt(n-1) == '.')
                leadOrTrailingZeros = true;

            boolean sPresent = false, pPresent = false;

            for(int i= 0; i<n; i++) {
                if((s.charAt(i) == 's' && pPresent == true) || (s.charAt(i) == 'p' && sPresent == false)) {
                    ans = false;
                }
                if(s.charAt(i) == 's')
                    sPresent = true;
                else if(s.charAt(i ) == 'p')
                    pPresent = true;
            }
            //System.out.println(ans);

            if(ans == false && sPresent && pPresent) {
                System.out.println("NO");
                continue;
            }

            if(sPresent && pPresent && leadOrTrailingZeros)
                System.out.println("NO");
            else
                System.out.println("YES");
        }
    }
}

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

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

Think step by step.

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


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

public class codechef {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static class FastWriter {
        private final BufferedWriter bw;

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

        public void print(Object o) throws IOException {
            bw.write(o.toString());
        }

        public void println(Object o) throws IOException {
            bw.write(o.toString());
            bw.newLine();
        }

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

    public static void main(String[] args) throws IOException {
        FastReader in = new FastReader();
        FastWriter out = new FastWriter();

        int t = in.nextInt();
//        int t=1;
        while (t-- > 0) {
            solve(in, out);
        }
        out.close();
    }

    public static void solve(FastReader in, FastWriter out) throws IOException {
        int n=in.nextInt();
        String s=in.next();
        int p=0;
        for(int i=0;i<s.length();i++){
            if(s.charAt(i)=='p'){
                p=i;
                break;
            }
        }
        for(int i=p+1;i<s.length();i++){
            if(s.charAt(i)=='s'){
                System.out.println("NO");
                return;
            }
        }

        if(s.length()>1 && s.contains("sp") && !((s.charAt(0)=='s' && s.charAt(1)=='p') ||
                (s.charAt(s.length()-1-1)=='s' && s.charAt(s.length()-1)=='p'))){
            System.out.println("NO");
            return;
        }
        System.out.println("Yes");
    }
}

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

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

Think step by step.

Response:
```plaintext
3
4
s.sp
6
pss..s
8
pspspsps
```