**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Guess the Maximum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob came up with a rather strange game. They have an array of integers $a_1, a_2,\ldots, a_n$. Alice chooses a certain integer $k$ and tells it to Bob, then the following happens:

  * Bob chooses two integers $i$ and $j$ ($1 \le i < j \le n$), and then finds the maximum among the integers $a_i, a_{i + 1},\ldots, a_j$;    * If the obtained maximum is strictly greater than $k$, Alice wins, otherwise Bob wins.

Help Alice find the maximum $k$ at which she is guaranteed to win.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^4$) — the number of elements in the array.

The second line of each test case contains $n$ integers $a_1, a_2,\ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^4$.

Output: For each test case, output one integer — the maximum integer $k$ at which Alice is guaranteed to win.

Examples:
input:
6
4
2 4 1 7
5
1 2 3 4 5
2
1 1
3
37 8 16
5
10 10 10 10 9
10
3 12 9 5 2 3 2 9 8 2
output:
3
1
0
15
9
2

Note: In the first test case, all possible subsegments that Bob can choose look as follows: $[2, 4], [2, 4, 1], [2, 4, 1, 7], [4, 1], [4, 1, 7], [1, 7]$. The maximums on the subsegments are respectively equal to $4, 4, 7, 4, 7, 7$. It can be shown that $3$ is the largest integer such that any of the maximums will be strictly greater than it.

In the third test case, the only segment that Bob can choose is $[1, 1]$. So the answer is $0$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int t,n,k,num[50005],maxn,bag[10005][5005];
bool cmp(int a,int b)
{
    return a<b;
}
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>num[i];
        }
        maxn=INT_MAX;
        for(int i=1;i<n;i++)
        {
            bag[i][i+1]=max(num[i],num[i+1]);
            maxn=min(maxn,bag[i][i+1]);
            for(int j=i+2;j<=n;j++)
            {
                bag[i][j]=max(bag[i][j-1],num[j]);
                maxn=min(maxn,bag[i][j]);
            }
        }
        cout<<maxn-1<<endl;
    }
}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases with the smallest and largest possible values of `n` and `a_i`.
2. **Special Patterns**: Arrays with all elements the same, arrays with elements in increasing or decreasing order, and arrays with a specific pattern that might cause the code to fail.
3. **Critical Values**: Values that might cause the minimum value to be incorrectly calculated, such as when the maximum of all subsegments is very close to the maximum element in the array.

Given the constraints and the nature of the problem, let's construct a test case that can expose the bug in the code. The bug in the code is that it calculates the minimum of the maximums of all subsegments and then subtracts 1, which might not always be the correct approach.

### Test Case 1: All Elements the Same
This test case checks if the code can handle arrays where all elements are the same.

```plaintext
1
5
3 3 3 3 3
```

### Test Case 2: Elements in Increasing Order
This test case checks if the code can handle arrays where elements are in increasing order.

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

### Test Case 3: Elements in Decreasing Order
This test case checks if the code can handle arrays where elements are in decreasing order.

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

### Test Case 4: Elements with a Specific Pattern
This test case checks if the code can handle arrays with a specific pattern that might cause the minimum value to be incorrectly calculated.

```plaintext
1
5
1 10 1 10 1
```

### Test Case 5: Maximum Element at the End
This test case checks if the code can handle arrays where the maximum element is at the end.

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

### Test Case 6: Maximum Element at the Beginning
This test case checks if the code can handle arrays where the maximum element is at the beginning.

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

### Test Case 7: Maximum Element in the Middle
This test case checks if the code can handle arrays where the maximum element is in the middle.

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

### Test Case 8: Large Array with a Single Maximum
This test case checks if the code can handle a large array with a single maximum element.

```plaintext
1
10
1 1 1 1 1 1 1 1 1 10
```

### Test Case 9: Large Array with All Elements Different
This test case checks if the code can handle a large array with all elements different.

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

### Test Case 10: Large Array with a Pattern
This test case checks if the code can handle a large array with a specific pattern.

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

### Combined Test Input
Combining all these test cases into a single input:

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

This combined test input should help expose any issues in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Guess the Maximum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob came up with a rather strange game. They have an array of integers $a_1, a_2,\ldots, a_n$. Alice chooses a certain integer $k$ and tells it to Bob, then the following happens:

  * Bob chooses two integers $i$ and $j$ ($1 \le i < j \le n$), and then finds the maximum among the integers $a_i, a_{i + 1},\ldots, a_j$;    * If the obtained maximum is strictly greater than $k$, Alice wins, otherwise Bob wins.

Help Alice find the maximum $k$ at which she is guaranteed to win.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^4$) — the number of elements in the array.

The second line of each test case contains $n$ integers $a_1, a_2,\ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^4$.

Output: For each test case, output one integer — the maximum integer $k$ at which Alice is guaranteed to win.

Examples:
input:
6
4
2 4 1 7
5
1 2 3 4 5
2
1 1
3
37 8 16
5
10 10 10 10 9
10
3 12 9 5 2 3 2 9 8 2
output:
3
1
0
15
9
2

Note: In the first test case, all possible subsegments that Bob can choose look as follows: $[2, 4], [2, 4, 1], [2, 4, 1, 7], [4, 1], [4, 1, 7], [1, 7]$. The maximums on the subsegments are respectively equal to $4, 4, 7, 4, 7, 7$. It can be shown that $3$ is the largest integer such that any of the maximums will be strictly greater than it.

In the third test case, the only segment that Bob can choose is $[1, 1]$. So the answer is $0$.

Buggy Code:
t=int(input())
for i in range(t):
    c=float('inf')
    n=int(input())
    s=list(map(int,input().split()))
    for j in range(len(s)-1):
        for k in range(j+1,len(s)):
            x=s[j:k+1]

            c=min(max(x),c)
    print(c-1)

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

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

Think step by step.

Response:
```plaintext
6
4
2 4 1 7
5
1 2 3 4 5
2
1 1
3
37 8 16
5
10 10 10 10 9
10
3 12 9 5 2 3 2 9 8 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Guess the Maximum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob came up with a rather strange game. They have an array of integers $a_1, a_2,\ldots, a_n$. Alice chooses a certain integer $k$ and tells it to Bob, then the following happens:

  * Bob chooses two integers $i$ and $j$ ($1 \le i < j \le n$), and then finds the maximum among the integers $a_i, a_{i + 1},\ldots, a_j$;    * If the obtained maximum is strictly greater than $k$, Alice wins, otherwise Bob wins.

Help Alice find the maximum $k$ at which she is guaranteed to win.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^4$) — the number of elements in the array.

The second line of each test case contains $n$ integers $a_1, a_2,\ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^4$.

Output: For each test case, output one integer — the maximum integer $k$ at which Alice is guaranteed to win.

Examples:
input:
6
4
2 4 1 7
5
1 2 3 4 5
2
1 1
3
37 8 16
5
10 10 10 10 9
10
3 12 9 5 2 3 2 9 8 2
output:
3
1
0
15
9
2

Note: In the first test case, all possible subsegments that Bob can choose look as follows: $[2, 4], [2, 4, 1], [2, 4, 1, 7], [4, 1], [4, 1, 7], [1, 7]$. The maximums on the subsegments are respectively equal to $4, 4, 7, 4, 7, 7$. It can be shown that $3$ is the largest integer such that any of the maximums will be strictly greater than it.

In the third test case, the only segment that Bob can choose is $[1, 1]$. So the answer is $0$.

Buggy Code:
import sys
# from collections import defaultdict
def input(): return sys.stdin.readline().strip()

t = int(input())
for _ in range(t):
    n = int(input())
    arr = list(map(int, input().split()))
    m = float('inf')
    for i in range(n):
        for j in range(i+2, n+1):
            tmp = max(arr[i:j])
            m = min(tmp, m)
    print(m-1)

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

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

Think step by step.

Response:
```plaintext
6
4
2 4 1 7
5
1 2 3 4 5
2
1 1
3
37 8 16
5
10 10 10 10 9
10
3 12 9 5 2 3 2 9 8 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Guess the Maximum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob came up with a rather strange game. They have an array of integers $a_1, a_2,\ldots, a_n$. Alice chooses a certain integer $k$ and tells it to Bob, then the following happens:

  * Bob chooses two integers $i$ and $j$ ($1 \le i < j \le n$), and then finds the maximum among the integers $a_i, a_{i + 1},\ldots, a_j$;    * If the obtained maximum is strictly greater than $k$, Alice wins, otherwise Bob wins.

Help Alice find the maximum $k$ at which she is guaranteed to win.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^4$) — the number of elements in the array.

The second line of each test case contains $n$ integers $a_1, a_2,\ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^4$.

Output: For each test case, output one integer — the maximum integer $k$ at which Alice is guaranteed to win.

Examples:
input:
6
4
2 4 1 7
5
1 2 3 4 5
2
1 1
3
37 8 16
5
10 10 10 10 9
10
3 12 9 5 2 3 2 9 8 2
output:
3
1
0
15
9
2

Note: In the first test case, all possible subsegments that Bob can choose look as follows: $[2, 4], [2, 4, 1], [2, 4, 1, 7], [4, 1], [4, 1, 7], [1, 7]$. The maximums on the subsegments are respectively equal to $4, 4, 7, 4, 7, 7$. It can be shown that $3$ is the largest integer such that any of the maximums will be strictly greater than it.

In the third test case, the only segment that Bob can choose is $[1, 1]$. So the answer is $0$.

Buggy Code:
import java.util.*;

public class guess_the_max
{
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      int t = sc.nextInt();
      for(int i=0;i<t;i++)
      {
         int n = sc.nextInt();
         int arr[] = new int[n];
         for(int j=0;j<n;j++)
         {
            arr[j] = sc.nextInt();
         }
         ArrayList <Integer> list = new ArrayList<>();
         for(int j=0;j<n;j++)
         {
            ArrayList <Integer> listin = new ArrayList<>();
            int count =0;
                  for(int k=j;k<n;k++)
                  {
                       if (count==0)
                       {
                           listin.add(arr[k]);
                           count++;
                       }
                       else
                       {
                        listin.add(arr[k]);
                        int max= Collections.max(listin);
                        list.add(max);
                       }

                  }

                }
                int max = Collections.min(list);
                System.out.println(max-1);

      }
   }
}

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

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

Think step by step.

Response:
```plaintext
3
5
5 4 3 2 1
2
10 10
4
1 100 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Guess the Maximum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob came up with a rather strange game. They have an array of integers $a_1, a_2,\ldots, a_n$. Alice chooses a certain integer $k$ and tells it to Bob, then the following happens:

  * Bob chooses two integers $i$ and $j$ ($1 \le i < j \le n$), and then finds the maximum among the integers $a_i, a_{i + 1},\ldots, a_j$;    * If the obtained maximum is strictly greater than $k$, Alice wins, otherwise Bob wins.

Help Alice find the maximum $k$ at which she is guaranteed to win.

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

The first line of each test case contains a single integer $n$ ($2 \le n \le 5 \cdot 10^4$) — the number of elements in the array.

The second line of each test case contains $n$ integers $a_1, a_2,\ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^4$.

Output: For each test case, output one integer — the maximum integer $k$ at which Alice is guaranteed to win.

Examples:
input:
6
4
2 4 1 7
5
1 2 3 4 5
2
1 1
3
37 8 16
5
10 10 10 10 9
10
3 12 9 5 2 3 2 9 8 2
output:
3
1
0
15
9
2

Note: In the first test case, all possible subsegments that Bob can choose look as follows: $[2, 4], [2, 4, 1], [2, 4, 1, 7], [4, 1], [4, 1, 7], [1, 7]$. The maximums on the subsegments are respectively equal to $4, 4, 7, 4, 7, 7$. It can be shown that $3$ is the largest integer such that any of the maximums will be strictly greater than it.

In the third test case, the only segment that Bob can choose is $[1, 1]$. So the answer is $0$.

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

public class find {
    public static int findSmallest(int[] arr) {
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < arr.length; i++) {
            for (int j = i + 1; j < arr.length; j++) {
                int[] arr1 = Arrays.copyOfRange(arr, i, j + 1);
                min = Math.min(getMaxKey(arr1), min);
            }
        }
        return min - 1;
    }

    public static int getMaxKey(int[] arr) {
        int max = Integer.MIN_VALUE;
        for (int j : arr) {
            max = Math.max(j, max);
        }
        return max;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        while (n > 0) {
            int m = scanner.nextInt();
            int[] arr = new int[m];
            for (int i = 0; i < m; i++) {
                arr[i] = scanner.nextInt();
            }
            System.out.printf("%d\n", findSmallest(arr));
            --n;
        }
    }
}

**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
6
4
2 4 1 7
5
1 2 3 4 5
2
1 1
3
37 8 16
5
10 10 10 10 9
10
3 12 9 5 2 3 2 9 8 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
#include<iostream> #include<cmath>#include<algorithm>#include<string>#include<iomanip>#include<vector>#include <math.h>#include <bits/stdc++.h>
using namespace std;
int main() {

int t;
cin>>t;

char c;
int n,P,B;

while(t--)
{
      cin>>c>>n;
      if(c=='P')
            {P=n;}
      else
      {
            if(P>=n)
            {
                  P=P-n;
                  cout<<"NO"<<endl;
            }
            else
            {     P=0;
                  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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's logic fails. The primary issue in the code is that it does not correctly handle the number of people at the bus stop when a bus arrives. Specifically, it does not properly account for the number of people who remain at the bus stop after a bus leaves.

Let's break down the steps to create a test input:

1. **Initial Setup**: Start with a small number of people at the bus stop.
2. **Bus Arrival**: Introduce a bus with a small number of seats to see if the code correctly handles the situation where all people can board the bus.
3. **People Arrival**: Add more people to the bus stop to create a scenario where the next bus cannot accommodate all the people.
4. **Multiple Buses**: Introduce buses with different seat capacities to test the code's ability to handle multiple buses and the remaining people at the bus stop.
5. **Edge Cases**: Include edge cases such as a bus with exactly enough seats for all people at the bus stop, and a bus with one more seat than the number of people at the bus stop.

Here is the test input that follows these steps:

```plaintext
10
P 2
B 2
P 3
B 4
P 5
B 5
P 6
B 10
P 1
B 1
P 1
B 1
```

### Explanation:
1. **P 2**: 2 people arrive at the bus stop.
2. **B 2**: A bus with 2 seats arrives. All 2 people board the bus. Monocarp can board the bus (output: YES).
3. **P 3**: 3 more people arrive at the bus stop.
4. **B 4**: A bus with 4 seats arrives. All 3 people board the bus, and there is 1 seat left for Monocarp (output: YES).
5. **P 5**: 5 more people arrive at the bus stop.
6. **B 5**: A bus with 5 seats arrives. All 5 people board the bus, and there are no seats left for Monocarp (output: NO).
7. **P 6**: 6 more people arrive at the bus stop.
8. **B 10**: A bus with 10 seats arrives. All 6 people board the bus, and there are 4 seats left for Monocarp (output: YES).
9. **P 1**: 1 more person arrives at the bus stop.
10. **B 1**: A bus with 1 seat arrives. The person boards the bus, and there are no seats left for Monocarp (output: NO).
11. **P 1**: 1 more person arrives at the bus stop.
12. **B 1**: A bus with 1 seat arrives. The person boards the bus, and there are no seats left for Monocarp (output: NO).

This test input will help expose the bug in the provided code by testing various scenarios where the number of people and bus seats are critical.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

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

int main()
{
    int t;
    cin >> t;
    int p = 0;

    while (t--)
    {
        int b = 0;
        int n;
        char c;
        cin >> c >> n;
        if (c == 'P')
        {
            p = n;
        }
        else
        {
            b = n;
            if (b > p)
            {
                cout << "YES" << endl;
                p = 0;
            }
            else
            {
                cout << "NO" << endl;
                p -= b;
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
10
P 1
P 2
B 3
P 1
B 2
P 1
B 1
P 1
B 1
B 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
#include <iostream>
#include<string>
using namespace std;
int  main(int argc, const char** argv) {
    int B=0 ,P=0,t=0;
    int p;
    int count ;
    cin>>count;
    for (int i = 1; i <= count; i++)
    {
         char c;
    cin>>c ;

     if (c=='P')
    {
        cin>>p;
        p+=P;
    }
    else if (c=='B')
    { int b;
        cin>>b;
        b+=B;
        t=p-b;
        if (t>=0)
        {
            cout<<"NO\n";
            P+=p;
            B+=b;
        }
        else{
        cout<<"YES\n";
        B =0 ;
             P=0;
        }

    }

    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The primary issues in the code are:

1. Incorrect handling of the number of people at the bus stop.
2. Incorrect logic for determining if Monocarp can board the bus.

### Step-by-Step Analysis:

1. **Incorrect Handling of People at the Bus Stop:**
   - The code incorrectly accumulates the number of people at the bus stop in the variable `P` and the number of free seats in the variable `B`.
   - It does not correctly reset the number of people at the bus stop after a bus leaves.

2. **Incorrect Logic for Determining Monocarp's Boarding:**
   - The code checks if the number of people at the bus stop is greater than or equal to the number of free seats, and if so, it prints "NO" and incorrectly updates the number of people and free seats.
   - It should instead check if there are enough seats for all people (except Monocarp) and then determine if Monocarp can board the bus.

### Test Input to Expose the Bug:

We need a test input that:
- Causes the code to accumulate an incorrect number of people at the bus stop.
- Results in an incorrect determination of whether Monocarp can board the bus.

### Example Test Input:

```plaintext
6
P 1
B 1
P 1
B 2
P 1
B 1
```

### Explanation:

1. **Event 1: P 1**
   - 1 person arrives at the bus stop.
   - People at the stop: 1

2. **Event 2: B 1**
   - A bus with 1 free seat arrives.
   - The bus can take all 1 person, and Monocarp can board the bus.
   - Expected output: "YES"
   - Correct behavior: People at the stop: 0

3. **Event 3: P 1**
   - 1 person arrives at the bus stop.
   - People at the stop: 1

4. **Event 4: B 2**
   - A bus with 2 free seats arrives.
   - The bus can take all 1 person, and Monocarp can board the bus.
   - Expected output: "YES"
   - Correct behavior: People at the stop: 0

5. **Event 5: P 1**
   - 1 person arrives at the bus stop.
   - People at the stop: 1

6. **Event 6: B 1**
   - A bus with 1 free seat arrives.
   - The bus can take all 1 person, but Monocarp cannot board the bus.
   - Expected output: "NO"
   - Correct behavior: People at the stop: 0

### Expected Output:
```plaintext
YES
YES
NO
```

This test input should expose the bug in the buggy code by showing incorrect handling of people at the bus stop and incorrect determination of Monocarp's boarding.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
def monocarp_bus(events):
    people_at_stop = 0
    result = []

    for event in events:
        if event[0] == 'P':
            people_at_stop += int(event[1])
        elif event[0] == 'B':
            available_seats = int(event[1])
            if people_at_stop < available_seats:
                result.append("YES")
                people_at_stop = 0
            else:
                result.append("NO")
                people_at_stop -= available_seats

    return result

# Example usage:
n = 10
events = [
    ('P', 2),
    ('P', 5),
    ('B', 8),
    ('P', 14),
    ('B', 5),
    ('B', 9),
    ('B', 3),
    ('P', 2),
    ('B', 1),
    ('B', 2)
]

print("\n".join(monocarp_bus(events)))

**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
P 10
B 10
B 9
P 1
B 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
n, ch, sp = int(input()), 0, []
for i in range(n):
    str = input()
    if str[0] == 'P':
        ch += int(str[2:])
    if str[0] == 'B':
        if (int(str[2]) > ch):
            ch = 0
            sp.append(1)
        else:
            ch -= int(str[2:])
            sp.append(0)
for j in range(len(sp)):
    if sp[j] == 1:
        print('YES')
    else:
        print('NO')

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

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

Think step by step.

Response:
```plaintext
10
P 1
P 1
P 1
B 3
P 1
B 2
P 1
B 2
P 1
B 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
n=int(input())
l=[]
for i in range (n):
    f=input().split(" ")
    l.append(f[0]+f[1])
people=0
ans=[]
for k in l:
    if k[0]=="P":
        people=people+int(k[1:])
    else:
        if(people<int(k[1])):
            people=0
            ans.append("YES")
        elif(people>=int(k[1])):
            people=people-int(k[1])
            ans.append("NO")
for l in ans:
    print(l)

**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
P 1
B 1
P 1
B 1
B 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
import java.util.Scanner ;

public class App {

    public static void main(String[] args) {

        Scanner read = new Scanner(System.in);

        int n = read.nextInt() ;

        int p = 0 , b = 0;

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

            char smp = read.next().charAt(0);

            int q = read.nextInt();

            if ( smp == 'P' ){
                p=0;
                p+=q;
                b=0;
            }else{
                b+=q;
                if ( b > p ){
                    System.out.println("YES");
                }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
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;
public class Waiting_For {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        ArrayList<Character> s = new ArrayList<>();
        ArrayList<Integer> num = new ArrayList<>();

        int sum = 0;

        int number = sc.nextInt();

        for(int i=0; i<number; i++)
        {
            s.add(sc.next().charAt(0));
            num.add(sc.nextInt());
        }

        for(int i=0; i<number; i++)
        {
            if(s.get(i) == 'P')
            {
                sum += num.get(i);
            }
            else
            {
                sum = num.get(i) - sum;

                if(sum >= 1)
                {
                    System.out.println("YES");
                }
                else
                {
                    System.out.println("NO");
                }

                sum = Math.abs(sum);

                if(i<number-1 && s.get(i+1) == 'P')
                {
                    sum = 0;
                }
            }
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
10
P 1
P 2
B 3
P 1
B 2
P 1
B 1
P 1
B 2
P 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
#include<iostream>
#include<vector>
#include<string>
#include<math.h>
#include<map>
#include<set>
#include<algorithm>
#include<utility>
#include<limits>
#include<unordered_set>
#include<deque>
#define ll long long
#define pb push_back
#define vll vector<ll>
#define print(ans) cout<< ans <<endl
#define srt(v) sort(v.begin(), v.end())
#define VAGHANI ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
using namespace std;
const int MOD = 1e9 + 7;
const ll n = 100;
ll hsh[n];

//
//ll mode(ll base,ll k,ll mod){
//
//    ll result=1;
//    base = base%MOD;
//    while(k>0){
//
//        if(k%2==1){
//            result=(result  * base)%mod;
//        }
//
//        k = k >>1;
//        base = (base * base)%mod;
//
//    }
//    return result;
//}

int main()
{

    VAGHANI
    ll PS=0;
    ll FS=0;
    ll t;

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

        char c;
        ll n;
        cin>>c>>n;

        if(c=='P')
            PS+=n;
        else
            FS+=n;

        if(c=='B'){
            if(FS>=PS){
                FS-=PS;
                PS=0;
            }else{
                if(PS!=0){
                    PS=PS-FS;
                    FS=0;
                }

            }
        }
        if(c=='B'){
            if(FS>=1){
                FS--;
                PS++;
                cout<<"YES"<<endl;
            }else{

                cout<<"NO"<<endl;
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
10
P 1
P 2
B 3
P 1
B 1
P 1
B 1
P 1
B 1
B 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

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

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int tt; cin >> tt;
    int sum = 0;
    while(tt--){
        char o; cin >> o;
        int a; cin >> a;
        if (o=='P'){
            sum += a;
        } else if(o=='B'){
            sum -= a;
            cout << (sum<=0 ? "YES" : "NO") << endl;
            // cout << sum << endl;
            sum++;
        }
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
P=1
for i in range(int(input())):
    e, n = (input().split())
    if e=="P":
        P+=int(n)
    elif e=="B":
        if int(n)>=P:
            P=1
            print("YES")
        if int(n)<=P:
            P-=int(n)
            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
5
P 1
B 1
P 1
B 2
B 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

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

using namespace std;
#define ll long long

int main() {
    ll cnt = 0, n;
    cin >> n;
    char s;
    while (n) {
        ll a;
        cin >> s >> a;
        if (s == 'P') {
            cnt += a;
        }
        if (s == 'B') {
            if (cnt <= a) {
                cout << "YES" << endl;
            } else {
                cout << "NO" << endl;
            }
            cnt -= (a - 1);
            if (cnt < 0) {
                cnt = 0;
            }
        }
        n--;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The main issue in the buggy code is that it incorrectly handles the number of people who can board the bus. Specifically, it subtracts `(a - 1)` from the count of people, which can lead to incorrect results.

### Steps to Generate the Test Input:
1. **Initial Setup**: Start with a few events to set up the initial conditions.
2. **Edge Case 1**: Test a scenario where the bus has exactly enough seats for all the people at the stop.
3. **Edge Case 2**: Test a scenario where the bus has more seats than the people at the stop, but Monocarp can still board.
4. **Edge Case 3**: Test a scenario where the bus has fewer seats than the people at the stop, and Monocarp cannot board.
5. **Edge Case 4**: Test a scenario where the bus has more seats than the people at the stop, but Monocarp chooses not to board.

### Challenging Test Input:
```plaintext
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
```

### Explanation:
1. **Initial Setup**:
   - `P 2` and `P 5` add 7 people to the bus stop.
2. **Edge Case 1**:
   - `B 8` (8 seats): There are 7 people at the stop. Monocarp can board the bus.
3. **Edge Case 2**:
   - `P 14` adds 14 more people, making the total 21 people at the stop.
   - `B 5` (5 seats): Only 5 people can board, and Monocarp cannot board.
4. **Edge Case 3**:
   - `B 9` (9 seats): 16 people are left at the stop. 9 people can board, and Monocarp cannot board.
5. **Edge Case 4**:
   - `B 3` (3 seats): 7 people are left at the stop. 3 people can board, and Monocarp cannot board.
6. **Edge Case 5**:
   - `P 2` adds 2 more people, making the total 6 people at the stop.
   - `B 1` (1 seat): 6 people are left at the stop. 1 person can board, and Monocarp cannot board.
7. **Edge Case 6**:
   - `B 2` (2 seats): 5 people are left at the stop. 2 people can board, and Monocarp can board.

This test input should expose the bug in the buggy code by challenging the incorrect handling of the number of people who can board the bus.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
t,c,l=int(input()),0,0
for i in range(t):
  a,b=input().split()
  b=int(b)
  if a=='P':
    c+=b
  else:
    l+=b
  if a=='B':
    if c==0:
      c=d
    if l>c:
      print('YES')
    else:
      print('NO')
  if a=='P':
    l=0
  else:
    d=c
    c=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
10
P 1
P 2
B 3
P 1
B 1
B 2
P 3
B 5
B 1
B 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
import java.util.Scanner;

public class J_Waiting_for {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        sc.nextLine();
        int rem = 0;
        for (int i = 0; i < n; i++){
            String event = sc.nextLine();
            char type = event.charAt(0);
            int num = Integer.parseInt(event.substring(2));
            if (type == 'P'){
                rem += num;
            }else{
                rem = Math.max(0,rem - num);
                if (rem == 0){
                    System.out.println("YES");
                    rem = 1;
                }else{
                    System.out.println("NO");
                }
            }

        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
5
P 1
B 1
P 1
B 1
B 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main() {
    int n;
    cin >> n;

   vector<pair<char, int>> commands;

    for (int i = 0; i < n; ++i) {
        char command;
        int value;
       cin >> command >> value;
        commands.emplace_back(command, value);
    }
    int check = 0 ;
    int p = 0 ;
    for (auto num : commands) {
        if (num.first == 'P'){
            if (check != 0 || p < 0 ) {
                p = 0 ;
                check = 0 ;
            }
            p += num.second;

        }
        if (num.first == 'B') {
            p -= num.second;
            if (p < 0 ) {
                cout << "YES" << "\n" ;
            }
            if (p>= 0 ) {
                cout << "NO"  << "\n";
            }
            ++check ;
        }
    }

    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
5
P 10
B 5
P 2
B 10
B 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin>>t;
    int i=0;
    int count=0;
    while(i<t){
        int n;
        char a;
        while(true){
            cin>>a>>n;
            i++;
            if(a=='B'){
                break;
            }
            count+=n;
        }
        if(n>=(count+1)){
            cout<<"YES"<<endl;
            count=0;
        }
        else{
            cout<<"NO"<<endl;
            count=count-n;
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
10
P 1
P 1
P 1
B 2
P 1
B 1
B 1
P 1
B 2
B 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
import java.util.*;

public class Hello{
    public static void main(String[] args) {
        Scanner take = new Scanner(System.in);
        int n = take.nextInt();
        int ans=0;
        while(n > 0){
            char c = take.next().charAt(0);
            int k = take.nextInt();
            take.nextLine();
            if(c=='B'){
                if(ans<k){
                    System.out.println("Yes");
                }
                else
                {
                    System.out.println("No");
                }
                if(ans<n)
                ans=0;
                else ans-=k;
            }
            else if(c=='P')
            ans+=k;
            n--;
        }
    }
}

**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 buggy code, we need to focus on scenarios where the logic for handling the number of people and the number of seats available in the bus might fail. Specifically, the code has issues with correctly updating the number of people at the bus stop and determining if Monocarp can board the bus.

### Key Points to Consider:
1. **Initial People at the Bus Stop**: Ensure the initial number of people at the bus stop is correctly managed.
2. **Bus Capacity and People**: Check how the code handles the number of people trying to board the bus and the remaining seats.
3. **Monocarp's Decision**: Ensure the code correctly determines if Monocarp can board the bus based on the remaining seats.

### Test Input to Expose the Bug:
We will create a test input that includes:
- A bus arriving with fewer seats than the number of people at the stop.
- A bus arriving with exactly the number of seats as the number of people at the stop.
- A bus arriving with more seats than the number of people at the stop.
- A bus arriving with no people at the stop (Monocarp can always board).

### Example Test Input:
```plaintext
6
P 5
B 4
P 1
B 1
B 6
P 2
```

### Explanation:
1. **Initial State**: 0 people at the bus stop.
2. **Event 1**: 5 people arrive. Now, 5 people are at the bus stop.
3. **Event 2**: A bus with 4 seats arrives. 4 people board the bus, 1 person remains. Monocarp cannot board (4 < 5).
4. **Event 3**: 1 more person arrives. Now, 2 people are at the bus stop.
5. **Event 4**: A bus with 1 seat arrives. 1 person boards the bus, 1 person remains. Monocarp cannot board (1 < 2).
6. **Event 5**: A bus with 6 seats arrives. 1 person boards the bus, 0 people remain. Monocarp can board (6 > 1).
7. **Event 6**: 2 more people arrive. Now, 2 people are at the bus stop.

### Expected Output:
```plaintext
NO
NO
YES
```

This test input will help expose the bug in the buggy code, particularly in how it handles the number of people at the bus stop and the remaining seats on the bus.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n= in.nextInt(),Person,seat,TotalPersons=0,validation=0;
        char Char;
      while(n-->0){
          Char=in.next().charAt(0);
          if (Char =='P'){
              if(validation>0){
                  TotalPersons=0;
              }validation=0;
              Person =in.nextInt();
              TotalPersons+=Person;

          }
          if(Char=='B'){
              validation++;
              seat = in.nextInt();
              TotalPersons-=seat;
              if(TotalPersons>=0){
                  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
10
P 1
P 2
B 3
P 4
B 5
P 6
B 7
P 8
B 9
B 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

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

using namespace std;

int main(){
    int t;
    cin>>t;
    char prev = 'A';
    long long sum = 0;
    char ch[t];
    int a[t];

    for(int i=0;i<t;i++) cin>>ch[i]>>a[i];

    int idx = 0;
    while(idx < t){
        if(ch[idx] == 'P'){
            sum = 0;
        }
        while(idx < t and ch[idx] == 'P')
        {
            sum += a[idx];
            idx++;
        }

        while(idx < t and ch[idx] == 'B')
        {
            sum -= a[idx];
            if(sum >= 0) cout << "NO" << endl;
            else cout << "YES" << endl;
            idx++;
        }
    }
    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
5
P 1
P 1
B 2
P 1
B 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Waiting for...
time_limit_ms: 2000
memory_limit_mb: 512
Description: Monocarp is waiting for a bus at the bus stop. Unfortunately, there are many people who want to ride a bus too.

You are given a list of events of two types:

- B $$$b_i$$$ — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ — $$$p_i$$$ people arrive at the stop.

These events are listed in a chronological order.

When a bus arrives, the following happens. All people at the bus stop (except for Monocarp) try to enter the bus. If there are enough free seats for all of them, then they all enter the bus. Otherwise, some people remain at the bus stop (the number of people who enter the bus is equal to the number of free seats).

If there is still at least one free seat after all people (except for Monocarp) enter the bus, then Monocarp can decide to enter this bus as well (but he might choose to wait for another bus). For each bus, you have to determine if it is possible for Monocarp to take that bus.

Input: The first line contains one integer $$$n$$$ $$$(1 \le n \le 10^3)$$$ — the number of events.

Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains the description of the $$$i$$$-th event in one of the two following formats:

- B $$$b_i$$$ ($$$1 \le b_i \le 10^6$$$) — a bus with $$$b_i$$$ free seats arrives at the stop;
- P $$$p_i$$$ ($$$1 \le p_i \le 10^6$$$) — $$$p_i$$$ people arrive at the stop.

Additional constraint on the input: there is at least one event of type B.

Output: For each event of type B, print YES if it is possible for Monocarp to take the corresponding bus, or NO otherwise (case-insensitive).

Examples:
input:
10
P 2
P 5
B 8
P 14
B 5
B 9
B 3
P 2
B 1
B 2
output:
YES
NO
NO
YES
NO
YES

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
//643894
int main(){
    int t, n = 0, h;
    char g;
    cin >> h;
    while(h--){
        cin >> g >> t;
        if(g == 'P') n += t;
        if(g == 'B') {
            n -= t;
            if(n < 0) {
                n = 0;
                cout << "YES" << '\n';
            }
            else cout << "NO" << '\n';
        }
    }
    if(g == 'P'){
        if(n < 0) cout << "YES" << '\n';
        else cout << "NO" << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on the logic of how the code handles the number of people at the bus stop and the number of free seats in the bus. The bug in the code is that it incorrectly updates the number of people at the bus stop and checks if Monocarp can board the bus.

### Key Points to Consider:
1. **Incorrect Update of People at the Bus Stop:**
   - The code subtracts the number of free seats from the number of people at the bus stop (`n -= t`).
   - This is incorrect because it should first check if the number of people at the bus stop is less than or equal to the number of free seats, and only then subtract the number of people from the bus stop.

2. **Monocarp's Boarding Logic:**
   - The code incorrectly checks if `n < 0` to determine if Monocarp can board the bus.
   - This is incorrect because `n < 0` means there are negative people at the bus stop, which is not a valid scenario.

### Test Input:
We need a test input that highlights these issues. A good test input should:
- Have a bus with a small number of free seats arriving when there are more people at the bus stop.
- Have a bus with a large number of free seats arriving when there are fewer people at the bus stop.
- Ensure that the number of people at the bus stop is correctly updated after each bus event.

### Chosen Test Input:
```plaintext
5
P 10
B 5
P 2
B 8
B 3
```

### Explanation:
1. **Event 1: P 10**
   - 10 people arrive at the bus stop.
   - People at the bus stop: 10

2. **Event 2: B 5**
   - A bus with 5 free seats arrives.
   - 5 people board the bus.
   - People at the bus stop: 5
   - Monocarp cannot board the bus because there are no free seats left.
   - Expected output: NO

3. **Event 3: P 2**
   - 2 more people arrive at the bus stop.
   - People at the bus stop: 7

4. **Event 4: B 8**
   - A bus with 8 free seats arrives.
   - 7 people board the bus.
   - People at the bus stop: 0
   - Monocarp can board the bus because there is 1 free seat left.
   - Expected output: YES

5. **Event 5: B 3**
   - A bus with 3 free seats arrives.
   - No people are at the bus stop.
   - Monocarp can board the bus.
   - Expected output: YES

### Expected Output:
```plaintext
NO
YES
YES
```

This test input should expose the bug in the code by showing that the number of people at the bus stop is not correctly updated and that the logic for determining if Monocarp can board the bus is flawed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

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

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main() {
    int tc;
    cin>>tc;
    while(tc--){
        long long n;
        cin>>n;
        long long m;
        cin>>m;
        long long r;
        cin>>r;
        long long c;
        cin>>c;
        long long value=m-c+(max(n,m)+m-1)*(n-r);
        cout<<value<<endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code's logic might fail. The buggy code calculates the sum of Manhattan distances using the formula:

\[ \text{value} = m - c + (\max(n, m) + m - 1) \cdot (n - r) \]

This formula seems to be incorrect because it does not account for the actual movement of each person. Let's consider a few scenarios:

1. **Small Grid with a Middle Position**: This can help verify the basic movement logic.
2. **Large Grid with a Corner Position**: This can test the boundary conditions and the scaling of the formula.
3. **Single Row or Column**: This can test edge cases where the grid is degenerate.
4. **Middle of a Large Grid**: This can test the general case and the correctness of the movement logic.

Let's start with a small grid and a middle position to see if the formula holds:

### Test Case 1: Small Grid with Middle Position
- \( n = 3 \)
- \( m = 3 \)
- \( r = 2 \)
- \( c = 2 \)

This is a 3x3 grid, and the person at position (2, 2) leaves. The movements should be:
- Person 5 moves to (2, 2)
- Person 6 moves to (2, 3)
- Person 8 moves to (3, 2)
- Person 9 moves to (3, 3)

The distances are:
- Person 5: \( |1-2| + |2-2| = 1 \)
- Person 6: \( |1-2| + |3-2| = 2 \)
- Person 8: \( |2-2| + |2-3| = 1 \)
- Person 9: \( |2-3| + |3-3| = 1 \)

Sum of distances: \( 1 + 2 + 1 + 1 = 5 \)

### Test Case 2: Large Grid with Corner Position
- \( n = 1000 \)
- \( m = 1000 \)
- \( r = 1 \)
- \( c = 1 \)

This is a 1000x1000 grid, and the person at position (1, 1) leaves. The movements should be:
- Person 2 moves to (1, 1)
- Person 3 moves to (1, 2)
- Person 4 moves to (1, 3)
- ...
- Person 1000000 moves to (1000, 1000)

The distances are:
- Person 2: \( |1-1| + |2-1| = 1 \)
- Person 3: \( |1-1| + |3-1| = 2 \)
- ...
- Person 1000000: \( |1000-1| + |1000-1| = 1998 \)

Sum of distances: \( 1 + 2 + 3 + ... + 1998 \)

Using the formula for the sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{1998} k = \frac{1998 \cdot 1999}{2} = 1997001 \]

### Test Case 3: Single Row
- \( n = 1 \)
- \( m = 100 \)
- \( r = 1 \)
- \( c = 50 \)

This is a 1x100 grid, and the person at position (1, 50) leaves. The movements should be:
- Person 51 moves to (1, 50)
- Person 52 moves to (1, 51)
- ...
- Person 100 moves to (1, 100)

The distances are:
- Person 51: \( |1-1| + |51-50| = 1 \)
- Person 52: \( |1-1| + |52-51| = 1 \)
- ...
- Person 100: \( |1-1| + |100-99| = 1 \)

Sum of distances: \( 1 + 1 + 1 + ... + 1 = 50 \)

### Test Case 4: Middle of a Large Grid
- \( n = 500 \)
- \( m = 500 \)
- \( r = 250 \)
- \( c = 250 \)

This is a 500x500 grid, and the person at position (250, 250) leaves. The movements should be:
- Person 125001 moves to (250, 250)
- Person 125002 moves to (250, 251)
- ...
- Person 250000 moves to (500, 500)

The distances are:
- Person 125001: \( |250-250| + |251-250| = 1 \)
- Person 125002: \( |250-250| + |252-250| = 2 \)
- ...
- Person 250000: \( |500-250| + |500-250| = 500 \)

Sum of distances: \( 1 + 2 + 3 + ... + 500 \)

Using the formula for the sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{500} k = \frac{500 \cdot 501}{2} = 125250 \]

### Challenging Test Input
```plaintext
4
3 3 2 2
1000 1000 1 1
1 100 1 50
500 500 250 250
```

This test input includes a mix of small and large grids, and positions that are in the middle, corners, and single rows, which 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: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

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

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
#include <iostream>

using namespace std;

int main()
{
    long long t,n,m,r,c,h;

    cin>>t;
    for(int i=0; i<t; i++){
            cin>>n>>m>>r>>c;
       if(n==r)
       cout<<m-c<<"\n";
    else if(n==m&&r!=n&&c!=m){
           h=(n-r+m-c)*m;
        cout<<h<<"\n";
    }
        else if(n==m==r==c)
            cout<<"0\n";
        else if(n==r&&m==c)
            cout<<"0\n";
        else if(m==c&&n!=r)
            cout<<(n-r)*c+(c-1)*(n-r);
            else if(n!=r&&m==c)
                cout<<((n-r)*m)+(n-r+m-c)+((c-1)*(n-r))-1<<"\n";
        else
            cout<<((n-r)*m)+(n-r+m-c)+((c-1)*(n-r))<<"\n";

    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
10 10 10 10
```