**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

Examples:
input:
3
5
1 0
1 1
3 0
5 0
2 1
3
0 0
1 0
3 0
9
1 0
2 0
3 0
4 0
5 0
2 1
7 1
8 1
9 1
output:
4
0
8

Note: The four triangles in question for the first test case:

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    points = []
    for _ in range(n):
        x, y = map(int, input().split())
        points.append((x, y))

    dic = {}

    for val in points:
        if(val[0] in dic):
            dic[val[0]].append(val[1])

        else:
            dic[val[0]] = [val[1]]

    count = 0
    for k in dic.keys():
        if(len(dic[k]) == 2):
            count =count + (n-2)

        for val in dic[k]:
            if(val == 0):
                if((k+2,val) in points and (k+1,val+1) in points):
                    count = count + 1
            if(val == 1):
                if((k+2,val) in points and (k+1,val-1) in points):
                    count = count + 1

    print(count)

**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 test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has a few issues:

1. It incorrectly counts the number of right triangles.
2. It does not correctly handle the formation of right triangles using the points.

Let's construct a small-scale test input that can expose these issues:

### Test Case 1: Points forming a right triangle
We can create a set of points that form a clear right triangle, but the code might not correctly identify it.

### Test Case 2: Points forming multiple right triangles
We can create a set of points that form multiple right triangles to ensure the code correctly counts all possible triangles.

### Test Case 3: Points that do not form any right triangles
We can create a set of points that do not form any right triangles to ensure the code correctly handles this scenario.

### Test Case 4: Points with the same x-coordinate
We can create a set of points with the same x-coordinate to check how the code handles such cases.

### Test Case 5: Points with the same y-coordinate
We can create a set of points with the same y-coordinate to check how the code handles such cases.

### Test Case 6: Points with mixed y-coordinates
We can create a set of points with mixed y-coordinates to ensure the code correctly handles mixed scenarios.

Here is the test input:

```plaintext
6
4
0 0
1 1
2 0
3 1
3
0 0
1 0
2 0
5
0 0
1 0
2 0
0 1
1 1
4
0 0
1 0
2 1
3 1
3
0 0
0 1
1 0
5
0 0
1 0
2 0
0 1
1 1
```

### Explanation:
1. **Test Case 1**:
   - Points: (0, 0), (1, 1), (2, 0), (3, 1)
   - Expected Output: 2 (Right triangles: (0, 0), (1, 1), (2, 0) and (1, 1), (3, 1), (2, 0))

2. **Test Case 2**:
   - Points: (0, 0), (1, 0), (2, 0)
   - Expected Output: 0 (No right triangles)

3. **Test Case 3**:
   - Points: (0, 0), (1, 0), (2, 0), (0, 1), (1, 1)
   - Expected Output: 2 (Right triangles: (0, 0), (1, 0), (0, 1) and (1, 0), (2, 0), (1, 1))

4. **Test Case 4**:
   - Points: (0, 0), (1, 0), (2, 1), (3, 1)
   - Expected Output: 2 (Right triangles: (0, 0), (1, 0), (2, 1) and (1, 0), (3, 1), (2, 1))

5. **Test Case 5**:
   - Points: (0, 0), (0, 1), (1, 0)
   - Expected Output: 1 (Right triangle: (0, 0), (0, 1), (1, 0))

6. **Test Case 6**:
   - Points: (0, 0), (1, 0), (2, 0), (0, 1), (1, 1)
   - Expected Output: 2 (Right triangles: (0, 0), (1, 0), (0, 1) and (1, 0), (2, 0), (1, 1))

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

Examples:
input:
3
5
1 0
1 1
3 0
5 0
2 1
3
0 0
1 0
3 0
9
1 0
2 0
3 0
4 0
5 0
2 1
7 1
8 1
9 1
output:
4
0
8

Note: The four triangles in question for the first test case:

Buggy Code:
import java.util.*;
import java.io.*;
public class SatyamCounting {
	public static void main(String [] args) {
		try {
			System.setIn(new FileInputStream("input.txt"));
			System.setOut(new PrintStream(new FileOutputStream("output.txt")));
		} catch (Exception e) {
			System.err.println("Error");
		}

		Scanner sc = new Scanner(System.in);
		int t1 = sc.nextInt();
		while (t1-- > 0) {
			int n = sc.nextInt();
			List<Integer> top = new ArrayList<>();
			List<Integer> bot = new ArrayList<>();
			for (int i = 0 ; i < n ; i++) {
				int a = sc.nextInt();
				int b = sc.nextInt();
				if (b == 1) {
					top.add(a);
				} else {
					bot.add(a);
				}
			}

			int tot = top.size() + bot.size();

			long ans = 0;
			for (int i : top) {
				if (bot.contains(i)) {
					ans += (tot - 2);
				}
				if (bot.contains(i - 1) && bot.contains(i + 1)) {
					ans++;
				}
			}

			for (int i : bot) {
				if (top.contains(i - 1) && top.contains(i + 1)) {
					ans++;
				}
			}

			System.out.println(ans);

		}
		sc.close();
	}
	static int[] arrayGen(int size , int limit) {
		int [] ans = new int[size];
		for (int i = 0 ; i < size ; i++) {
			ans[i] = (int)(Math.random() * limit);
		}
		return ans;
	}

	static List<Integer> seive(int N) {
		boolean[] sieve = new boolean[N];

		for (int i = 0; i < sieve.length; i++) {
			sieve[i] = true;
		}
		for (int i = 2; i < Math.sqrt(N); i++) {
			if (sieve[i] == true) {
				for (int j = (i * i); j < N; j = j + i) {
					sieve[j] = false;
				}
			}
		}
		List<Integer> ls = new ArrayList<>();
		for (int i = 2; i < sieve.length; i++) {
			if (sieve[i] == true) {
				ls.add(i);
			}
		}
		return ls;
	}

	public static void swapInArray(int ind1 , int ind2 , int[] arr) {
		int temp = arr[ind1];
		arr[ind1] = arr[ind2];
		arr[ind2] = temp;
	}

	public static void printArray(int [] a) {
		for (int i : a) {
			System.out.print(i + " ");
		}
		System.out.println();
	}

	public static void printList(List<Integer> ls) {
		for (Integer i : ls) {
			System.out.print(i + " ");
		}
		System.out.println();
	}
	public static int gcd(int a , int b) {
		if (b == 0) {
			return a;
		}
		return gcd(b , a % 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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code seems to have issues with correctly counting the number of right triangles, especially when there are points with the same x-coordinate but different y-coordinates.

### Step-by-Step Analysis:
1. **Identify the Bug**: The code incorrectly counts right triangles by checking if a point exists in both the `top` and `bot` lists. It also has a condition that checks for points at `i-1` and `i+1` in the opposite list, which is not a general solution for all cases.

2. **Edge Cases**:
   - **Points with the same x-coordinate but different y-coordinates**: This can lead to incorrect counting of right triangles.
   - **Points with consecutive x-coordinates**: This can also lead to incorrect counting.

3. **Constructing the Test Input**:
   - We need a test case with points that have the same x-coordinate but different y-coordinates.
   - We also need points with consecutive x-coordinates to test the `i-1` and `i+1` conditions.

### Challenging Test Input:
```plaintext
1
7
1 0
1 1
2 0
2 1
3 0
3 1
4 0
```

### Explanation:
- **Test Case 1**:
  - Points: (1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1), (4, 0)
  - **Expected Output**: 12
  - **Reasoning**:
    - Points (1, 0) and (1, 1) can form right triangles with (2, 0) and (2, 1).
    - Points (2, 0) and (2, 1) can form right triangles with (3, 0) and (3, 1).
    - Points (3, 0) and (3, 1) can form right triangles with (4, 0).
    - Each pair of points with the same x-coordinate can form right triangles with each point on the next x-coordinate, leading to a total of 12 right triangles.

This test input should expose the bug in the provided code by showing that it incorrectly counts the number of right triangles.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

Examples:
input:
3
5
1 0
1 1
3 0
5 0
2 1
3
0 0
1 0
3 0
9
1 0
2 0
3 0
4 0
5 0
2 1
7 1
8 1
9 1
output:
4
0
8

Note: The four triangles in question for the first test case:

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

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

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

    private static void solve() throws IOException
    {
        int n = sc.nextInt();
        boolean[][] arr = new boolean[ 2 ][ 200050 ];
        int max = 0;
        for( int i = 1; i <= n; i++ )
        {
            int x = sc.nextInt();
            int y = sc.nextInt();
            arr[ y ][ x ] = true;
            max = Math.max( max, x );
        }

        int cnt = 0;

        for( int i = 0; i <= max; i++ )
        {
            if( arr[ 0 ][ i ] && arr[ 1 ][ i ] ) cnt += n - 2;
            if( arr[ 0 ][ i ] && arr[ 0 ][ i + 2 ] && arr[ 1 ][ i + 1 ] ) cnt++;
            if( arr[ 1 ][ i ] && arr[ 1 ][ i + 2 ] && arr[ 0 ][ i + 1 ] ) cnt++;
        }

        System.out.println( cnt );
    }
}

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

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

Think step by step.

Response:
To create a test input that exposes the bug in the given code, we need to focus on the logic that counts the number of right triangles. The buggy code has the following issues:

1. It incorrectly counts the number of right triangles by checking specific patterns.
2. It does not correctly handle all possible configurations of points that can form right triangles.

To expose these issues, we can create a test case with points that should form right triangles but are not counted correctly by the buggy code. Here's a step-by-step approach to generate such a test case:

1. **Points on the Same X-coordinate:**
   - The code checks if there are points at `(x, 0)` and `(x, 1)` and counts them as potential right triangles. However, it does not correctly count all possible right triangles formed by these points.

2. **Points on Different X-coordinates:**
   - The code checks specific patterns like `(x, 0)`, `(x+2, 0)`, and `(x+1, 1)` to form a right triangle. We need to create a test case where these points are present but the code fails to count all possible right triangles.

3. **Edge Cases:**
   - Test cases with the minimum number of points (3) and the maximum number of points (200,000).

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

```plaintext
1
5
1 0
2 0
3 0
1 1
2 1
```

Explanation:
- This test case has 5 points.
- Points at `(1, 0)` and `(1, 1)` can form a right triangle.
- Points at `(2, 0)` and `(2, 1)` can form another right triangle.
- Points at `(1, 0)`, `(2, 0)`, and `(2, 1)` can form a right triangle.
- Points at `(1, 1)`, `(2, 1)`, and `(3, 0)` can form a right triangle.

The correct number of right triangles is 4, but the buggy code might not count all of them correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

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

using namespace std;

typedef long long ll;

void solve() {
    int n, m;
    scanf("%d %d", &n, &m);
    vector<int> counts(7, 0);
    char inp[60];
    scanf("%s", inp);
    while (n--) {
        ++counts[inp[n-1]-'A'];
    }
    int ans = 0;
    for (int i = 0; i < 7; ++i) {
        ans += max(0, m-counts[i]);
    }
    printf("%d\n", ans);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    scanf("%d", &t);
    while (t--) solve();

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 1
A
7 1
ABCDEFG
10 2
AAABBBCCC
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
#include <iostream>
#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
#include <math.h>
#include <stdio.h>
#include <vector>
#define ll long long

int main(){
    ll t;
    cin>>t;
    while(t--){
        ll n,m;
        cin>>n>>m;
        string s;
        cin>>s;
        vector<ll> v(n,0);
        for(ll i=0;i<n;i++){
            if(s[i]=='A'){
                v[0]++;
            }
            if(s[i]=='B'){
                v[1]++;
            }
            if(s[i]=='C'){
                v[2]++;
            }
            if(s[i]=='D'){
                v[3]++;
            }
            if(s[i]=='E'){
                v[4]++;
            }
            if(s[i]=='F'){
                v[5]++;
            }
            if(s[i]=='G'){
                v[6]++;
            }
        }
        ll sum=0;
        for(ll i=0;i<7;i++){
            if(v[i]<m){
                sum+=m-v[i];
            }
        }
        cout<<sum<<endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that might cause the code to fail. Here are some key points to consider:

1. **Minimum and Maximum Values for \( n \) and \( m \)**: Test the code with the smallest and largest possible values for \( n \) and \( m \).
2. **Imbalanced Distribution of Problems**: Test with a string where certain difficulty levels are missing or are in very low numbers.
3. **Exact Number of Problems Needed**: Test with a string where the number of problems for each difficulty level is exactly what is needed for the rounds.
4. **Single Round Scenarios**: Test with \( m = 1 \) to see if the code handles the simplest case correctly.
5. **Multiple Rounds Scenarios**: Test with \( m > 1 \) to see if the code handles multiple rounds correctly.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
3
5 1
ABCDF
10 2
AABBCDDEFF
5 5
AABCD
```

### Explanation:
1. **Test Case 1:**
   - \( n = 5 \), \( m = 1 \)
   - String: `ABCDF`
   - This test case checks if the code can handle a scenario where all difficulty levels except 'G' are present, and only one round is needed.
   - Expected output: 1 (since one 'G' is missing)

2. **Test Case 2:**
   - \( n = 10 \), \( m = 2 \)
   - String: `AABBCDDEFF`
   - This test case checks if the code can handle a scenario where there are enough problems for one round but not enough for two rounds.
   - Expected output: 2 (since two 'A's, two 'B's, and two 'G's are needed, and only one 'A' and one 'B' are missing)

3. **Test Case 3:**
   - \( n = 5 \), \( m = 5 \)
   - String: `AABCD`
   - This test case checks if the code can handle a scenario where there are not enough problems for the required rounds.
   - Expected output: 20 (since each round needs 7 problems, and there are only 5 problems in total, with significant gaps in the required difficulties)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        int a,b;
        cin>>a>>b;
        string s;
        cin>>s;
        unordered_map<char,int>mp;
        for(int i=0;i<s.length();i++){
            mp[s[i]]++;
        }
        int ans=0;
        for(auto i:mp){
            if(i.second<b) ans=b-i.second;
        }
        if(mp.size()<7){
            ans+=(7-mp.size())*b;
        }
        cout<<ans<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 5
AAAAA
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
t=int(input())
for i in range(t):
    n, m = map(int, input().split())
    s=input()
    d={}
    lst=[]
    for i in s:
        lst.append(i)
    for i in lst:
        if i in d:
            d[i]=d[i]+1
        else:
            d[i]=1
    sample=['A','B','C','D','E','F','G']
    cd=0
    if m>=2:
        for i in d:
            if d[i]==1:
                cd+=1
        count=0
        for i in sample:
            if i not in lst:
              count+=1
        final=cd+count+count
        print(final)
    else:
        count=0
        for i in sample:
            if i not in lst:
                count+=1
        final=count
        print(final)

**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
7 2
ABCDEFG
7 2
ABCDFEG
7 2
ABCDGFE
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
t=int(input())
for i in range(t):
    n,m=map(int,input().split())
    s=str(input())
    ans=0
    if('A' not in s):
        ans+=1
    if('B' not in s):
        ans+=1
    if('C' not in s):
        ans+=1
    if('D' not in s):
        ans+=1
    if('E' not in s):
        ans+=1
    if('F' not in s):
        ans+=1
    if('G' not in s):
        ans+=1
    if(m==1):
        print(ans)
    elif(m==2):
        print(ans*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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider the following:

1. **Edge Cases**: The code incorrectly handles cases where `m > 1`. Specifically, it adds an extra 1 to the result when `m == 2`, which is not always correct.
2. **Scenarios**: We should create a scenario where the bug is most evident, i.e., where the code's output is significantly different from the correct output.

Let's break it down:

- **Test Case 1**: `m = 1` and the string has all difficulties except 'A' and 'F'. The correct output should be 2, and the buggy code should also output 2.
- **Test Case 2**: `m = 2` and the string has all difficulties except 'A' and 'F'. The correct output should be 4, but the buggy code will output 5.
- **Test Case 3**: `m = 2` and the string has all difficulties except 'A', 'F', and 'G'. The correct output should be 6, but the buggy code will output 7.

Here is the test input:

```plaintext
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 2
BBCDEFFGG
```

Explanation:
- **Test Case 1**: `n = 10`, `m = 1`, `s = "BGECDCBDED"`. The string lacks 'A' and 'F', so the correct output is 2. The buggy code will also output 2.
- **Test Case 2**: `n = 10`, `m = 2`, `s = "BGECDCBDED"`. The string lacks 'A' and 'F', so the correct output is 4. The buggy code will output 5.
- **Test Case 3**: `n = 9`, `m = 2`, `s = "BBCDEFFGG"`. The string lacks 'A', 'F', and 'G', so the correct output is 6. The buggy code will output 7.

This test input will effectively expose the bug in the code.

3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 2
BBCDEFFGG


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
def find_min_problems(t, test_cases):
    results = []
    for i in range(t):
        n, m = test_cases[i][:2]
        a = test_cases[i][2]
        count = {chr(c): 0 for c in range(ord('A'), ord('H'))}
        for char in a:
            count[char] += 1
        needed = 0
        for difficulty in count:
            required = m
            if count[difficulty] < required:
                needed += required - count[difficulty]
        results.append(needed)
    return results
t = 3
test_cases = [
    (10, 1, 'BGECDCBDED'),
    (10, 2, 'BGECDCBDED'),
    (9, 1, 'BBCDEFFGG')
]
results = find_min_problems(t, test_cases)
for result in results:
    print(result)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code might fail. The primary issue in the buggy code is that it incorrectly calculates the number of problems needed for each difficulty level. Specifically, it always requires `m` problems for each difficulty, even if there are already enough problems in the bank.

Here are the steps to create a challenging test input:

1. **Edge Case with No Problems in the Bank:**
   - If there are no problems in the bank, the code should require `7 * m` problems (one for each difficulty level 'A' to 'G').

2. **Edge Case with Few Problems in the Bank:**
   - If there are very few problems in the bank, the code should require a significant number of additional problems to meet the requirement.

3. **Edge Case with Just Enough Problems in the Bank:**
   - If there are exactly enough problems in the bank to meet the requirement, the code should not require any additional problems.

4. **Edge Case with More Than Enough Problems in the Bank:**
   - If there are more than enough problems in the bank for some difficulty levels, the code should not require additional problems for those levels.

Based on these considerations, here is a challenging test input:

```plaintext
4
0 2
5 1
ABCDG
7 1
ABCDEFG
10 1
AABBBCCDDE
```

### Explanation:
1. **Test Case 1:**
   - `n = 0`, `m = 2`
   - No problems in the bank.
   - The code should require `7 * 2 = 14` problems.

2. **Test Case 2:**
   - `n = 5`, `m = 1`
   - Problems in the bank: 'ABCDG'
   - The code should require `7 - 5 = 2` problems (one for 'E' and one for 'F').

3. **Test Case 3:**
   - `n = 7`, `m = 1`
   - Problems in the bank: 'ABCDEFG'
   - The code should require `0` problems (since there are exactly enough problems for one round).

4. **Test Case 4:**
   - `n = 10`, `m = 1`
   - Problems in the bank: 'AABBBCCDDE'
   - The code should require `1` problem (one for 'F').

This test input will help expose the bug in the buggy code by covering various edge cases and ensuring that the correct number of additional problems is calculated.

4
0 2
5 1
ABCDG
7 1
ABCDEFG
10 1
AABBBCCDDE


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

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

public class Main {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int t= in.nextInt();
        int a=0;
        int b=0;
        int c=0;
        int d=0;
        int e=0;
        int f=0;
        int p=0;
        int y=0;
        for (int i = 0; i < t; i++) {
            int n= in.nextInt();
            int m= in.nextInt();
            String g= in.next();
            for (int j = 0; j < n; j++) {
                if (g.charAt(j)=='A'){
                    a++;
                }
                if (g.charAt(j)=='B'){
                    b++;
                }
                if (g.charAt(j)=='c'){
                    c++;
                }
                if (g.charAt(j)=='D'){
                    d++;
                }
                if (g.charAt(j)=='E'){
                    e++;
                }
                if (g.charAt(j)=='F'){
                    f++;
                }
                if (g.charAt(j)=='G'){
                    p++;
                }

            }
           if (a<m){
               y+=(m-a);
           }
           if (b<m){
               y+=(m-b);
           }
            if (c<m){
                y+=(m-c);
            }
            if (d<m){
                y+=(m-d);
            }
            if (e<m){
                y+=(m-e);
            }
            if (f<m){
                y+=(m-f);
            }
            if (p<m){
                y+=(m-p);
            }
            System.out.println(y-1); y=0;
        }

    }
}

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

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

Think step by step.

Response:
```plaintext
3
5 2
ABCDG
10 1
ABCDEFGGGG
7 3
ABCDEFF
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
/* package codechef; // don't place package name! */

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

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0)
		{
		    int n=sc.nextInt();
		    int m=sc.nextInt();
		    String s=sc.next();

		    int[] freq = new int[7];  // 'A' to 'G' mapped to indices 0 to 6
            for (char ch : s.toCharArray()) {
                freq[ch - 'A']++;
            }
            int fullRounds = Integer.MAX_VALUE;
            for (int count : freq) {
                fullRounds = Math.min(fullRounds, count);
            }
            int newProblemsNeeded = 0;
            for (int i = 0; i < 7; i++) {
                int remainingRounds = m - fullRounds;
                if (remainingRounds > 0 && freq[i] < remainingRounds) {
                    newProblemsNeeded += (remainingRounds - freq[i]);
                }
            }

            System.out.println(newProblemsNeeded);
        }

	}
}

**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
7 2
ABCDEFG
10 1
AAAAAAABCD
14 3
GGGGGGGGGGGGGG
```

3
7 2
ABCDEFG
10 1
AAAAAAABCD
14 3
GGGGGGGGGGGGGG


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

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

int main() {
    int t;
    cin>>t;
    while(t--) {
        int n, m;
        cin>>n>>m;
        string s;
        cin>>s;
        int ques = 7;
        unordered_map<char, int> mp;
        for(auto x : s) {
            mp[x]++;
        }
        int num = 0;
        for(int i=0;i<s.size();i++) {
            if(mp.find(s[i]) != mp.end()) {
                num += (m > mp[s[i]]) ? (m - mp[s[i]]) : 0;
            }
        }
        cout<<(num + ((7 - mp.size()) *m))<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
7 1
ABCDEFG
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

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

int main()
{
    int t ;
    cin >> t;
    while (t--){
    int n ,  m;
    char s[n+5] ;
    int counter =0;
    cin >> n >> m;
    for (int i = 0 ; i< n;i++)
        cin >> s[i];

    for (int i =0 ; i < 7 ; i++)
    {
        if (count (s , s + n ,char(65 + i))  < m )
            counter += m - count (s , s + n ,char(65 + i));
    }
    cout << counter<<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 consider the following points:

1. **Edge Cases**: The code should handle the minimum and maximum values of \( n \) and \( m \) correctly.
2. **Character Distribution**: The code should correctly count the number of each difficulty level ('A' to 'G') and determine the minimum number of problems needed.
3. **Input Length**: The input should be small-scale but still challenging.

Let's consider a test case where the input is such that the code might fail to correctly count the number of problems needed. Specifically, we can create a case where the number of problems of each difficulty level is just below the required number for \( m \) rounds.

### Test Input
```plaintext
1
5 2
ABCDF
```

### Explanation
- **Test Case Count**: 1
- **Number of Problems \( n \)**: 5
- **Number of Rounds \( m \)**: 2
- **Problems String**: "ABCDF"

### Expected Output
- For \( m = 2 \) rounds, we need 2 problems of each difficulty level ('A' to 'G').
- The input string "ABCDF" has:
  - 1 'A'
  - 1 'B'
  - 1 'C'
  - 1 'D'
  - 1 'F'
  - 0 'E'
  - 0 'G'

To hold 2 rounds, we need:
- 1 more 'A'
- 1 more 'B'
- 1 more 'C'
- 1 more 'D'
- 2 'E'
- 2 'F'
- 2 'G'

Total additional problems needed: \( 1 + 1 + 1 + 1 + 2 + 1 + 2 = 9 \)

### Expected Output
```plaintext
9
```

This test input should expose the bug in the code by requiring a significant number of additional problems to meet the requirements for 2 rounds.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
a=int(input())
for i in range(a):
    n,k=map(int,input().split())
    s=input()
    l=[]
    cou=1
    tot=0
    sum=[]
    for j in s:
        l.append(j)
    l.sort()
    for j in range(len(l)-1):
        if(l[j]==l[j+1]):
            cou+=1
        else:
            sum.append(l[j])
            if(cou<k):
                tot+=y-cou
            cou=1
    sum.append(l[len(l)-1])
    if(cou<k):
        tot+=k-cou
    print(tot+k*(7-len(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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The buggy code has several issues, including incorrect handling of the required number of problems for each difficulty level and incorrect counting of the total number of problems needed.

### Key Points to Consider:
1. **Edge Cases:**
   - Very few problems in the bank.
   - All problems in the bank are of the same difficulty.
   - The number of problems in the bank is just enough to cover the rounds.
   - The number of problems in the bank is not enough to cover the rounds.

2. **Scenarios:**
   - When `m` is 1.
   - When `m` is greater than 1.
   - When the bank has a mix of difficulties, but some difficulties are missing.

### Chosen Test Input:
Let's choose a test input where the number of problems in the bank is just enough to cover the rounds, but one difficulty is missing. This will expose the bug in the code's logic for counting the required problems.

```plaintext
3
7 1
ABCDEFZ
14 2
ABCDEFZABCDEFZ
15 3
ABCDEFZABCDEFZG
```

### Explanation:
1. **Test Case 1:**
   - `n = 7`, `m = 1`
   - `a = "ABCDEFZ"`
   - This test case has exactly one problem for each difficulty level, so Vlad needs to come up with one more problem of any difficulty to complete the round.

2. **Test Case 2:**
   - `n = 14`, `m = 2`
   - `a = "ABCDEFZABCDEFZ"`
   - This test case has exactly two problems for each difficulty level, so Vlad needs to come up with two more problems of any difficulty to complete the rounds.

3. **Test Case 3:**
   - `n = 15`, `m = 3`
   - `a = "ABCDEFZABCDEFZG"`
   - This test case has exactly three problems for each difficulty level except 'G', which has only one problem. Vlad needs to come up with two more problems of difficulty 'G' to complete the rounds.

These test cases are designed to expose the bug in the code's logic for counting the required problems and handling the required number of problems for each difficulty level.

3
7 1
ABCDEFZ
14 2
ABCDEFZABCDEFZ
15 3
ABCDEFZABCDEFZG


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
#!/usr/bin/python3

import collections

num = int(input())

for i in range(num):
    n, m = map(int, input().split())

    probs = [a for a in input()]

    counter = dict()
    max_letter = 'A'
    max_count = 0
    for char in probs:
        counter[char] = counter.get(char, 0) + 1
    for key in counter:
        if counter[key] > max_count:
            max_letter = key
            max_count = counter[key]
    max_count = min(max_count, m)
    res = 0

    req_probs = ['A', 'B', 'C', 'D', 'E', 'F', 'G']
    for req in req_probs:
        res += max(max_count - counter.get(req, 0), 0)

    print(res)

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

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

Think step by step.

Response:
```plaintext
1
7 5
AAAAAAA
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

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

        int t = scan.nextInt();

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

            int len = scan.nextInt();

            int round = scan.nextInt();

            scan.nextLine();

            String s = scan.nextLine();

            Map<Character , Integer> count = new HashMap<>();

            count.put('A' , 0);
            count.put('C' , 0);
            count.put('D' , 0);
            count.put('E' , 0);
            count.put('F' , 0);
            count.put('G' , 0);

            for (int j = 0; j < len; j++) {
                if (s.charAt(j) == 'A'){
                    count.merge('A' , 1 , (integer, integer2) -> integer + 1);
                }
                else if (s.charAt(j) == 'B'){
                    count.merge('B' , 1 , (integer, integer2) -> integer + 1);
                }

                else if (s.charAt(j) == 'C'){
                    count.merge('C' , 1 , (integer, integer2) -> integer + 1);
                }
                else if (s.charAt(j) == 'D'){
                    count.merge('D' , 1 , (integer, integer2) -> integer + 1);
                }
                else if (s.charAt(j) == 'E'){
                    count.merge('E' , 1 , (integer, integer2) -> integer + 1);
                }
                else if (s.charAt(j) == 'F'){
                    count.merge('F' , 1 , (integer, integer2) -> integer + 1);
                }
                else if (s.charAt(j) == 'G'){
                    count.merge('G' , 1 , (integer, integer2) -> integer + 1);
                }
            }

            int answer = 0;

            for (Map.Entry<Character , Integer> k : count.entrySet()){
                if (k.getValue() < round){
                    answer += round - k.getValue();
                }
            }

            System.out.println(answer);

        }
    }
}

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

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

Think step by step.

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

1. **Missing Initialization**: The map `count` is missing the initialization for the character 'B', which means any problems of difficulty 'B' will not be counted.
2. **Edge Cases**: We should consider edge cases where the number of problems of a certain difficulty is zero or very close to the required number of rounds.
3. **Small Scale**: The test input should be small enough to be manageable and within the constraints.

Given these points, let's construct a test case that will expose the missing initialization for 'B':

### Test Case:
- **Number of test cases**: 1
- **Number of problems**: 7
- **Number of rounds**: 1
- **Difficulty levels**: 'B' 'A' 'C' 'D' 'E' 'F' 'G'

This test case ensures that there is exactly one problem of each difficulty level, except for 'B', which is missing from the map initialization. The expected output should be 1 because one more problem of difficulty 'B' is needed to hold the round.

### Test Input:
```plaintext
1
7 1
BACDEFG
```

### Explanation:
- The input specifies one test case.
- There are 7 problems in the bank, and 1 round is to be held.
- The problems have difficulty levels 'B', 'A', 'C', 'D', 'E', 'F', 'G'.
- The buggy code will fail to count the 'B' problems correctly because 'B' is not initialized in the map.
- The expected output is 1, but the buggy code will output 0 because it does not account for the 'B' problems.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
import java.util.*;
public class Main{
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            int n = sc.nextInt();
            int m = sc.nextInt();
            sc.nextLine(); // Consume the newline after the integers
            String s = sc.nextLine();

            Map<Character, Integer> mp = new HashMap<>();
            mp.put('A', 0);
            mp.put('B', 0);
            mp.put('C', 0);
            mp.put('D', 0);
            mp.put('E', 0);
            mp.put('F', 0);

            // Count occurrences of each character in the string
            for (int i = 0; i < s.length(); i++) {
                char ch = s.charAt(i);
                mp.put(ch, mp.getOrDefault(ch, 0) + 1);
            }

            int count = 0;
            // For each character in the map, check if its count is less than m
            for (char ch : mp.keySet()) {
                if (mp.get(ch) < m) {
                    count += (m - mp.get(ch));
                }
            }

            System.out.println(count);
        }
    }
}

**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
50 5
ABCDEFGABCDEFGABCDEFGABCDEFGABCDEFG
10 1
GGGGGGGGGG
1 5
A
```

3
50 5
ABCDEFGABCDEFGABCDEFGABCDEFGABCDEFG
10 1
GGGGGGGGGG
1 5
A


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
#include <bits/stdc++.h>
#define FIO ios::sync_with_stdio(false);cin.tie(NULL)
#define debug(x) cout<<#x<<' '<<x<<endl;
#define  mod (long long)10e9+7
using namespace std;
#define all(x) x.begin(),x.end()
void NO(){cout<<"NO"<<endl;}
void YES(){cout<<"YES"<<endl;}
#define fr(i, x, n) for ( int i = x; i < n; i++)
#define frr(i, x, n) for (int i = x; i >= n; i--)

#define pn cout<<"No"<<endl
#define py cout<<"Yes"<<endl
#define pN cout<<"NO"<<endl
#define pY cout<<"YES"<<endl
#define pb push_back
#define sz(x) (int)(x).size()
#define mp make_pair
#define ff first
#define ss second
#define lb lower_bound
#define acc accumulate
#define ub upper_bound
#define ins insert
#define lb lower_bound
#define up upper_bound

typedef long long ll;

#define VI vector<int>
#define VLL vector<long long>

bool isprime(int n){
	if(n<2)	return 0;

   for(int i=2;i*i<=n;i++){
     if(n%i==0){
       return 0;
     }
   }
   return true;
 }

 bool issorted(string &v,int size){
 	for(int i=1;i<size;i++){
 		if(v[i]<v[i-1]) return 0;
 	}
 	return 1;
 }

 bool cmp(pair<int,int> &a,pair<int,int> &b){
 	return (a.ss<b.ss);

 }
template <typename T> void print( vector<T> v,ll size){ for(int i=0;i<size;i++){	cout<<v[i]<<' ';	}cout<<endl;}
template <typename T> void input( vector<T> &v,ll size){for(auto &it:v) cin>>it;}

 ll lcm(int a,int b){
 	return (a*b/(__gcd(a,b)));
 }

bool isperfectsq(int n){
	double p=sqrt(n);
	int k=(long)p;
	return (p-k>0)?0:1;
}

// vector<int > v(100100);
// 	fill(all(v),1);
// 	v[0]=0,v[1]=0;
// 	for(int i=0;i<=100050;i++){
// 		if(v[i]==1){
// 			for(int j=i+i;j<=100050;j+=i){
// 				v[j]=0;
// 			}
// 		}
// 	}
// 	vector< int > prime;
// 	for(int i=0;i<100050;i++){
// 		if(v[i]==1){
// 			prime.pb(i);
// 		}
// 	}

void solve(){
	int n,k;cin>>n>>k;
	string s;cin>>s;

	unordered_map<char,int> m;
	for(char i='A';i<='F';i++){
		m[i]=0;
	}
	fr(i,0,sz(s)){
		m[s[i]]++;
	}
	ll ans=0;
	for(auto it:m){
		if(it.ss<k){
			ans+=(k-it.ss);
		}
	}
	cout<<ans<<endl;

}

int32_t main(){
  FIO;
 #ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

 int t=1;
 cin>>t;

for(int i=1;i<=t;i++){
	// debug(i)
	solve();
 }

}
/*
  A-Z 65-90 to convert to small +32
  a-z 97-122 to convert to capital -32
  0-9 48-57
*/

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

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

Think step by step.

Response:
```plaintext
1
7 1
AABCDEF
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main(){
    int tc; cin >> tc;
    cin.ignore();
    while(tc--){
        int n,m; cin >> n >> m;
        string s; cin >> s;
        map<char,int> mp;
        for(int i = 0;i < n;i++){
            mp[s[i]]++;
        }
        int dem = 0;
        string a = {'A','B','C','D','E','F','G'};
        for(int i = 0;i < a.length();i++){
            if(s.find(a[i]) == string :: npos){
                dem += m;
            }
            else{
                if(mp[s[i]] < m){
                    dem += m - mp[s[i]];
                }
            }
        }
        cout << dem << endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
7 2
ABCDEFG
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
def min_problems_to_create(test_cases):
    results = []

    for n, m, problems in test_cases:
        # Initialize a dictionary to keep track of counts of each difficulty
        difficulty_count = {chr(65 + i): 0 for i in range(7)}  # A to G

        # Populate the dictionary with the counts from the problems string
        for problem in problems:
            difficulty_count[problem] += 1

        # Calculate the number of problems Vlad needs to create
        additional_problems_needed = 0
        for difficulty in difficulty_count:
            additional_problems_needed += max(m - difficulty_count[difficulty], 0)

        results.append(additional_problems_needed)

    return results

# Example usage
test_cases = [
    (10, 1, "BGECDCBDED"),
    (10, 2, "BGECDCBDED"),
    (9, 1, "BBCDEFFGG")
]

results = min_problems_to_create(test_cases)

for result in results:
    print(result)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, let's analyze the problem and the code to identify potential edge cases and scenarios that could cause issues.

### Analysis:
1. **Initialization of `difficulty_count`:**
   - The dictionary is correctly initialized to count the occurrences of each difficulty level from 'A' to 'G'.

2. **Population of `difficulty_count`:**
   - This part seems correct as it counts the occurrences of each difficulty level in the given problems string.

3. **Calculation of `additional_problems_needed`:**
   - The code calculates the number of additional problems needed for each difficulty level by using `max(m - difficulty_count[difficulty], 0)`.
   - This approach is incorrect because it does not account for the fact that Vlad needs to have at least `m` problems for each difficulty level in total, not just the difference.

### Challenging Test Input:
To expose this bug, we need a test case where the number of problems for one or more difficulty levels is less than `m`, but the total number of problems is enough to cover the required rounds. This will show that the current implementation incorrectly calculates the additional problems needed.

### Example Test Case:
- **Number of test cases (`t`):** 1
- **Number of problems (`n`):** 7
- **Number of rounds (`m`):** 2
- **Problems string (`a`):** "AABBBCC"

### Explanation:
- For `m = 2`, Vlad needs 2 problems of each difficulty level.
- The given problems string "AABBBCC" has:
  - 2 problems of difficulty 'A'
  - 3 problems of difficulty 'B'
  - 2 problems of difficulty 'C'
  - 0 problems of difficulty 'D'
  - 0 problems of difficulty 'E'
  - 0 problems of difficulty 'F'
  - 0 problems of difficulty 'G'

The correct number of additional problems needed is:
- 2 (for 'D')
- 2 (for 'E')
- 2 (for 'F')
- 2 (for 'G')

Total additional problems needed = 8.

However, the buggy code will calculate:
- `max(2 - 2, 0) = 0` for 'A'
- `max(2 - 3, 0) = 0` for 'B'
- `max(2 - 2, 0) = 0` for 'C'
- `max(2 - 0, 0) = 2` for 'D'
- `max(2 - 0, 0) = 2` for 'E'
- `max(2 - 0, 0) = 2` for 'F'
- `max(2 - 0, 0) = 2` for 'G'

Total additional problems needed (incorrectly calculated) = 8.

### Test Input:
```plaintext
1
7 2
AABBBCC
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
z=int(input())
for x in range(z):
    n,m=map(int,input().split(" "))
    s=input()
    r=0
    l=['A','B','C','D','E','F','G']
    for i in l:
        if i not in s:
            r=r+m
        else:
            c=0
            for j in s:
                if i==j:
                    c=c+1
            if(c<m):
                r=r+c
    print(r)

**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
ABCDE
5 5
BCDEF
7 5
ABCDEFG
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
import java.util.*;
public class ProblemGenerator{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0)
        {
            //System.out.println("enter n");
            int n=sc.nextInt();
            //System.out.println("enter m");
            int m=sc.nextInt();
            //System.out.println("enter s");
            String s=sc.next();
            Map<Character,Integer> mp=new HashMap<>();
            for(int i=0;i<n;i++)
            {
                mp.put(s.charAt(i),mp.getOrDefault(s.charAt(i),0)+1);
            }
            int ms=mp.size();
            int r=Math.abs(7-ms);
            int c=0;
            for(int i=0;i<ms;i++)
            {
                if(mp.containsKey(s.charAt(i)))
                {
                    if(mp.get(s.charAt(i))<m)
                 {
                    c+=m-mp.get(s.charAt(i));
                 }
                }
            }
            c+=r*m;
            System.out.println(c);
        }

    }
}

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

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

Think step by step.

Response:
```plaintext
1
7 5
AABBCCG
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem Generator
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vlad is planning to hold $m$ rounds next month. Each round should contain one problem of difficulty levels 'A', 'B', 'C', 'D', 'E', 'F', and 'G'.

Vlad already has a bank of $n$ problems, where the $i$-th problem has a difficulty level of $a_i$. There may not be enough of these problems, so he may have to come up with a few more problems.

Vlad wants to come up with as few problems as possible, so he asks you to find the minimum number of problems he needs to come up with in order to hold $m$ rounds.

For example, if $m=1$, $n = 10$, $a=$ 'BGECDCBDED', then he needs to come up with two problems: one of difficulty level 'A' and one of difficulty level 'F'.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 50$, $1 \le m \le 5$) — the number of problems in the bank and the number of upcoming rounds, respectively.

The second line of each test case contains a string $a$ of $n$ characters from 'A' to 'G' — the difficulties of the problems in the bank.

Output: For each test case, output a single integer — the minimum number of problems that need to come up with to hold $m$ rounds.

Examples:
input:
3
10 1
BGECDCBDED
10 2
BGECDCBDED
9 1
BBCDEFFGG
output:
2
5
1

Note:

Buggy Code:
import java.util.Scanner;

/**
 *
 * @author user
 */
public class PropA {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t= in.nextInt();

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

           int n=in.nextInt();
           int m=in.nextInt();
           String a=in.next();
           int A=0;
           int B=0;
           int C=0;
           int D=0;
           int E=0;
           int F=0;
           int G=0;
           int count=0;
        for(int j=0;j<m;j++){

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

             if(a.charAt(k)=='A'){
                A++;
             }
             if(a.charAt(k)=='B'){
               B++;
             }
             if(a.charAt(k)=='C'){
               C++;
             }
             if(a.charAt(k)=='D'){
                D++;
             }
             if(a.charAt(k)=='E'){
                 E++;
             }
             if(a.charAt(k)=='F'){
                 F++;
             }
             if(a.charAt(k)=='G'){
              G++;
             }

          }

        }
        int max= Math.max(A, Math.max(B, Math.max(C, Math.max(D, Math.max(E, Math.max(F, G))))));
            if(A==0){count=count+m;}else{
            if(A<(max/2)){
                count++;
             }}

            if(B==0){count=count+m;}else{
             if(B<(max/2)){
               count++;
             }}

            if(C==0){count=count+m;}else{
             if(C<(max/2)){
              count++;
             }}

            if(D==0){count=count+m;}else{
             if(D<(max/2)){
                count++;
             }}

            if(E==0){count=count+m;}else{
             if(E<(max/2)){
               count++;
             }}
            if(F==0){count=count+m;}else{
             if(F<(max/2)){
                  count++;
             }}
            if(G==0){count=count+m;}else{
             if(G<(max/2)){
               count++;
             }}

           System.out.println(count);
        }

    }

}

**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 test input that exposes the bug in the given buggy code, we need to focus on scenarios where the code's logic fails. The main issue in the buggy code is that it incorrectly calculates the number of problems needed for each difficulty level. Specifically, it checks if the count of each difficulty level is less than half of the maximum count, which is not the correct approach for determining the minimum number of problems needed.

### Steps to Generate the Test Input:
1. **Identify the Bug**: The bug is in the logic that checks if the count of each difficulty level is less than half of the maximum count. This is incorrect because the problem requires ensuring that there are enough problems of each difficulty level to cover all rounds, not just a comparison to half of the maximum count.

2. **Edge Cases**:
   - **Case 1**: All problems are of the same difficulty level.
   - **Case 2**: Some difficulty levels are completely missing.
   - **Case 3**: Uneven distribution of problems across difficulty levels.

3. **Construct the Test Input**:
   - **Case 1**: All problems are of difficulty 'A' and we need multiple rounds.
   - **Case 2**: No problems of difficulty 'G' and we need multiple rounds.
   - **Case 3**: Uneven distribution of problems across all difficulty levels.

### Chosen Test Input:
Let's choose a test case where the problems are unevenly distributed and some difficulty levels are missing. This will maximize the failure probability in the faulty solution.

```plaintext
3
5 3
AAAAA
10 2
ABCDEAAAAA
15 3
AABBCDDEEEFFFFG
```

### Explanation:
1. **Test Case 1**:
   - `n = 5`, `m = 3`
   - `a = "AAAAA"`
   - The code should output `14` because we need 3 problems of each difficulty level ('B', 'C', 'D', 'E', 'F', 'G') and we already have 5 problems of difficulty 'A'.

2. **Test Case 2**:
   - `n = 10`, `m = 2`
   - `a = "ABCDEAAAAA"`
   - The code should output `7` because we need 2 problems of each difficulty level, and we are missing 1 problem of difficulty 'F' and 1 problem of difficulty 'G'.

3. **Test Case 3**:
   - `n = 15`, `m = 3`
   - `a = "AABBCDDEEEFFFFG"`
   - The code should output `6` because we need 3 problems of each difficulty level, and we are missing 2 problems of difficulty 'A', 1 problem of difficulty 'B', 1 problem of difficulty 'C', and 1 problem of difficulty 'D'.

This test input will expose the bug in the logic of the buggy code by requiring a correct count of the minimum number of problems needed for each difficulty level.

3
5 3
AAAAA
10 2
ABCDEAAAAA
15 3
AABBCDDEEEFFFFG


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Perform Operations to Maximize Score
time_limit_ms: 3000
memory_limit_mb: 256
Description: I see satyam343. I'm shaking. Please more median problems this time. I love those. Please satyam343 we believe in you.

— satyam343's biggest fan

You are given an array $a$ of length $n$ and an integer $k$. You are also given a binary array $b$ of length $n$.

You can perform the following operation at most $k$ times:

  * Select an index $i$ ($1 \leq i \leq n$) such that $b_i = 1$. Set $a_i = a_i + 1$ (i.e., increase $a_i$ by $1$).

Your score is defined to be $\max\limits_{i = 1}^{n} \left( a_i + \operatorname{median}(c_i) \right)$, where $c_i$ denotes the array of length $n-1$ that you get by deleting $a_i$ from $a$. In other words, your score is the maximum value of $a_i + \operatorname{median}(c_i)$ over all $i$ from $1$ to $n$.

Find the maximum score that you can achieve if you perform the operations optimally.

For an arbitrary array $p$, $\operatorname{median}(p)$ is defined as the $\left\lfloor \frac{|p|+1}{2} \right\rfloor$-th smallest element of $p$. For example, $\operatorname{median} \left( [3,2,1,3] \right) = 2$ and $\operatorname{median} \left( [6,2,4,5,1] \right) = 4$.

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

Each test case begins with two integers $n$ and $k$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq k \leq 10^9$) — the length of the $a$ and the number of operations you can perform.

The following line contains $n$ space separated integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — denoting the array $a$.

The following line contains $n$ space separated integers $b_1, b_2, \ldots, b_n$ ($b_i$ is $0$ or $1$) — denoting the array $b$.

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

Output: For each test case, output the maximum value of score you can get on a new line.

Examples:
input:
8
2 10
3 3
1 1
3 10
3 3 3
0 0 0
4 4
2 1 5 1
0 1 0 1
5 4
7 5 2 5 4
0 0 1 0 1
5 1
5 15 15 2 11
1 0 0 1 1
5 2
10 11 4 10 15
1 1 0 1 0
4 4
1 1 2 5
1 1 0 0
2 1000000000
1000000000 1000000000
1 1
output:
16
6
8
13
21
26
8
3000000000

Note: For the first test case, it is optimal to perform $5$ operations on both elements so $a = [8,8]$. So, the maximum score we can achieve is $\max(8 + \operatorname{median}[8], 8 + \operatorname{median}[8]) = 16$, as $c_1 = [a_2] = [8]$. It can be proven that you cannot get a better score.

For the second test case, you are not able to perform operations on any elements, so $a$ remains $[3,3,3]$. So, the maximum score we can achieve is $3 + \operatorname{median}[3, 3] = 6$, as $c_1 = [a_2, a_3] = [3, 3]$. It can be proven that you cannot get a better score.

Buggy Code:
//To learn how to solve problems you need to solve problems.
#include<bits/stdc++.h>
#define int long long
using namespace std;
typedef long long LL;

const int N = 200010;
const LL inf = 1e17;

int T, n, k;

struct node{
    int val, c;
    bool operator <(const node u)const{
        return val < u.val;
    }
}a[N];

inline int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}

bool check(int mid)
{
    int cnt = 0, tar = (n - 1) / 2 + 1;
    for(int i = 1; i < n; i++) if(a[i].val >= mid) cnt++;

    //cout << cnt << endl;

    int rest = k;
    for(int i = n - 1; i && (cnt < tar); i--){
        if(a[i].val > mid || a[i].c == 0) continue;
        if(a[i].val + rest < mid) return false;
        rest -= mid - a[i].val, cnt++;
    }
    return (cnt >= tar);
}

void solve()
{
    n = read(), k = read();
    for(int i = 1; i <= n; i++) a[i].val = read();
    for(int i = 1; i <= n; i++) a[i].c = read();

    sort(a + 1, a + n + 1);

    int u = 0;
    for(int i = n; i; i--) if(a[i].c == 1){
        u = i;
        break;
    }

    int ans = 0;
    if(u){
        int pn = n / 2, x = 0, res = 0;
        for(int i = 1; i <= n; i++){
            if(i != u) x++;
            if(x == pn){
                res = a[i].val + a[u].val + k;
                break;
            }
        }
        ans = max(ans, res);
    }

    int res = a[n].val;

    //cout << check(5) << endl;

    int l = 1, r = 2000000000, mid, cur;
    while(l <= r){
        mid = (l + r) >> 1;
        if(check(mid)) cur = mid, l = mid + 1;
        else r = mid - 1;
    }
    res += cur;
    ans = max(ans, res);
    printf("%lld\n", ans);
}

signed main()
{
    //ios :: sync_with_stdio(false), cout.tie(0), cin.tie(0);
    cin >> T;
    while(T--) solve();
    return 0;
}
/*
1
5 1
5 15 15 2 11
1 0 0 1 1
*/

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

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

Think step by step.

Response:
```plaintext
1
3 2
1 2 3
0 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Perform Operations to Maximize Score
time_limit_ms: 3000
memory_limit_mb: 256
Description: I see satyam343. I'm shaking. Please more median problems this time. I love those. Please satyam343 we believe in you.

— satyam343's biggest fan

You are given an array $a$ of length $n$ and an integer $k$. You are also given a binary array $b$ of length $n$.

You can perform the following operation at most $k$ times:

  * Select an index $i$ ($1 \leq i \leq n$) such that $b_i = 1$. Set $a_i = a_i + 1$ (i.e., increase $a_i$ by $1$).

Your score is defined to be $\max\limits_{i = 1}^{n} \left( a_i + \operatorname{median}(c_i) \right)$, where $c_i$ denotes the array of length $n-1$ that you get by deleting $a_i$ from $a$. In other words, your score is the maximum value of $a_i + \operatorname{median}(c_i)$ over all $i$ from $1$ to $n$.

Find the maximum score that you can achieve if you perform the operations optimally.

For an arbitrary array $p$, $\operatorname{median}(p)$ is defined as the $\left\lfloor \frac{|p|+1}{2} \right\rfloor$-th smallest element of $p$. For example, $\operatorname{median} \left( [3,2,1,3] \right) = 2$ and $\operatorname{median} \left( [6,2,4,5,1] \right) = 4$.

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

Each test case begins with two integers $n$ and $k$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq k \leq 10^9$) — the length of the $a$ and the number of operations you can perform.

The following line contains $n$ space separated integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — denoting the array $a$.

The following line contains $n$ space separated integers $b_1, b_2, \ldots, b_n$ ($b_i$ is $0$ or $1$) — denoting the array $b$.

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

Output: For each test case, output the maximum value of score you can get on a new line.

Examples:
input:
8
2 10
3 3
1 1
3 10
3 3 3
0 0 0
4 4
2 1 5 1
0 1 0 1
5 4
7 5 2 5 4
0 0 1 0 1
5 1
5 15 15 2 11
1 0 0 1 1
5 2
10 11 4 10 15
1 1 0 1 0
4 4
1 1 2 5
1 1 0 0
2 1000000000
1000000000 1000000000
1 1
output:
16
6
8
13
21
26
8
3000000000

Note: For the first test case, it is optimal to perform $5$ operations on both elements so $a = [8,8]$. So, the maximum score we can achieve is $\max(8 + \operatorname{median}[8], 8 + \operatorname{median}[8]) = 16$, as $c_1 = [a_2] = [8]$. It can be proven that you cannot get a better score.

For the second test case, you are not able to perform operations on any elements, so $a$ remains $[3,3,3]$. So, the maximum score we can achieve is $3 + \operatorname{median}[3, 3] = 6$, as $c_1 = [a_2, a_3] = [3, 3]$. It can be proven that you cannot get a better score.

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

using namespace std;

typedef long long ll;

#define rep(i,a,b) for (ll i=a; i <=b; i++)
#define INF 1<<30
#define FOR(i,n) for(ll i=0; i<n; i++)
#define FORR(i,n) for(ll i=n-1; i>=0; i--)
#define pb push_back
#define vi vector<ll>

int main(){
    ll t; cin >> t;
    while (t--){
        ll n,k; cin >> n >> k;
        //n=rand()%10+2; k=rand()%100+1;
        pair<ll,ll> a[n];
        pair<ll,ll> b[n];
        FOR(i,n) cin >> a[i].first,a[i].second=i;
        FOR(i,n) cin >> b[i].first;
        /*
        FOR(i,n) a[i]={rand()%20+1,i};
        FOR(i,n) b[i].first=rand()%2;
        FOR(i,n) cout << a[i].first << " ";
        cout << endl;
        FOR(i,n) cout << b[i].first << " ";
        cout << endl;
        */
        multiset<ll> st;
        sort(a,a+n);
        ll e=1,it;
        FORR(i,n){
            if (e && b[a[i].second].first){
                e=0; it=i;
            }
            if (b[a[i].second].first && i!=n-1) st.insert(-a[i].first);
        }
        if (e) {
            cout << a[n-1].first+a[(n-1)/2].first << endl;
            continue;
        }
        ll mx=max(0ll,k+a[it].first-a[n-1].first+((it<=(n-2)/2 && (n-2)/2+1<n ? max(0ll,min(a[it].first+k, a[(n-2)/2+1].first)-a[(n-2)/2].first) : 0)));
        ll ps=(n-2)/2,nm=0,sm=0,cr=a[(n-2)/2].first;
        while (!st.empty() && ps+nm<n){
            auto pnt=st.lower_bound(-a[ps+nm].first);
            if (pnt==st.end()) break;
            if (k<a[ps+nm].first+*pnt+nm*(a[ps+nm].first-cr)) break;
            k-=a[ps+nm].first+*pnt+nm*(a[ps+nm].first-cr);
            sm+=a[ps+nm].first-cr;
            cr=a[ps+nm].first;
            st.erase(pnt);
            nm++;
        }
        if (ps+nm<n && nm) sm+=min(k/nm,a[ps+nm].first-cr);
        cout << a[n-1].first+a[ps].first+max(mx,sm) << endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 3
1 2 3 4 5
1 0 1 0 1
```