**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

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

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
from functools import lru_cache
import sys

def h(k, local_seq, threshold):
    seq = local_seq.copy()
    counts = [len(seq)] * (len(seq) + 1)

    for n in range(len(seq) + 1):
        seq_index = len(seq) - n

        if n % k == 0 and seq_index > 0:
            count_below_threshold = 0
        elif n <= k:
            count_below_threshold = 0
            for j in range(seq_index, len(seq)):
                if seq[j] <= threshold:
                    count_below_threshold += 1
        else:
            count_below_threshold = n
            for i in range((n-1)//k + 1):
                count_below_threshold = min(count_below_threshold, (1 if seq[seq_index+k*i] <= threshold else 0) + counts[seq_index + (k*i + 1)])
                if count_below_threshold == 0:
                    break

        counts[seq_index] = count_below_threshold

    return counts[0]

def best_median(k, input_seq):
    low = 0
    high = 10**9 + 1
    n = len(input_seq)

    final_len = k if n % k == 0 else n % k

    while high > low:
        med = low + (high - low)//2
        less_than_count = h(k, input_seq, med)

        #print("Binary search tracker", low, med, high, less_than_count)
        if less_than_count < (final_len + 1)//2:
            low = med+1
        else:
            high = med

    return low

def main():
    line_num = 0
    for line in sys.stdin:
        line = line.rstrip()
        if line_num == 0:
            num_examples = int(line)
        else:
            #print(line.split(" "))
            if line_num % 2 == 1:
                n, k = [int(x) for x in line.split(" ")]
            else:
                read_seq = [int(x) for x in line.split(" ")]
                assert len(read_seq) == n

                print(best_median(k, read_seq))
        line_num += 1
        if line_num == 2*num_examples + 1:
            break

if __name__ == "__main__":
    main()
    #print(h(3, [3, 9, 9, 2], 2))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. The buggy code uses a binary search approach to find the best median, but it may have issues with:

1. **Edge cases where the array length is exactly divisible by `k`**.
2. **Cases where the array has many repeated elements**.
3. **Cases where the array is already sorted or nearly sorted**.
4. **Cases where the array has a small number of elements**.

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

1. An array where the length is exactly divisible by `k`.
2. An array with many repeated elements.
3. An array that is already sorted.
4. A small array.

Here is a small-scale test input that includes these scenarios:

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

### Explanation:
1. **Test Case 1:**
   - `n = 6`, `k = 3`
   - Array: `[1, 2, 3, 4, 5, 6]`
   - The array length is exactly divisible by `k`. The expected output is `3` because the median of the remaining elements after removing subarrays of size 3 should be `3`.

2. **Test Case 2:**
   - `n = 5`, `k = 1`
   - Array: `[1, 1, 1, 1, 1]`
   - The array has many repeated elements. The expected output is `1` because all elements are the same.

3. **Test Case 3:**
   - `n = 4`, `k = 2`
   - Array: `[1, 2, 3, 4]`
   - The array is already sorted. The expected output is `2` because the median of the remaining elements after removing subarrays of size 2 should be `2`.

4. **Test Case 4:**
   - `n = 3`, `k = 3`
   - Array: `[1, 2, 3]`
   - The array length is exactly divisible by `k`. The expected output is `2` because the median of the remaining elements after removing the entire array should be `2`.

This test input should help expose any issues in the buggy code, particularly with handling edge cases and specific scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

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

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
problems = int(input())

def median(add, nums):

    bason = 0
    if len(nums) % 2 == 0:
        bason = 1
    return sorted([add[i] for i in nums])[len(nums)//2 - bason]

def generatePerms(cinx, n, left, pivot):

    if (left == 0 and cinx == n):
        return [[]]

    if cinx >= n:
        return []

    ans = set()

    for ea in generatePerms(cinx + pivot, n, left, pivot):
        ans.add(tuple(ea))
    for ea in range(cinx, n, pivot):

        for jth in generatePerms(ea + 1, n, left - 1, pivot):
            ans.add(tuple([cinx] + list(jth)))
    return ans

def solve():

    x, y = map(int, input().split(' '))

    basis = list(map(int, input().split(' ')))

    if x % y == 0:
        pierreTarjan = y
    else:
        pierreTarjan = x % y

    basils = generatePerms(0, x, pierreTarjan, y)
    ans = 0
    for ea in basils:
        ans = max(ans, median(basis, ea))
    print(ans)

for ea in range(problems):
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

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

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;

public class Prob12 {
	public static void main(String[] args) throws IOException {
		//setup
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer tokenizer = new StringTokenizer(in.readLine());
    	//reading integers
    	int t = Integer.parseInt(tokenizer.nextToken());

    	for(int i = 0; i < t; i++) {
    		//Use this everytime you are reading a new line
    		tokenizer = new StringTokenizer(in.readLine());
    		int n = Integer.parseInt(tokenizer.nextToken()), k = Integer.parseInt(tokenizer.nextToken());
    		tokenizer = new StringTokenizer(in.readLine());
    		long[] vals = new long[n];
    		for(int j = 0; j < n; j++) {
    			vals[j] = Long.parseLong(tokenizer.nextToken());
    		}
    		if(k >= n) {
    			Arrays.sort(vals);
    			System.out.println(vals[(n+1)/2 - 1]);
    			continue;
    		}
    		int size = n % k;
    		if(n % k == 0) size = k;
    		int need = size/2 + 1;
    		long lo = 0, hi = 1000000001;
//    		long lo = 0, hi = 20;
    		while (lo < hi) {
    			// find the middle of the current range (rounding up)
    			long mid = lo + (hi - lo + 1) / 2;
    			if (f(vals, mid, k, n, need, size - need)) {
    				// if mid works, then all numbers smaller than mid also work
    				lo = mid;
    			} else {
    				// if mid does not work, greater values would not work either
    				hi = mid - 1;
    			}
    		}
    		System.out.println(lo);
    	}

	}
	public static boolean f(long[] vals, long median, int k, int n, int remo, int remu) {
		if(remo <= 0) return true;
		if(remu < 0) return false;
		if(n <= 0) return false;
		boolean o = vals[n - 1] >= median;
		if(o) return f(vals, median, k, n - 1, remo - 1, remu) || f(vals, median, k, n - k, remo, remu);
		else return f(vals, median, k, n - 1, remo, remu - 1) || f(vals, median, k, n - k, remo, remu);
	}
//	public static boolean f(long[] vals, long median, int k, int n) {
//		int need = (n % k)/2 + 1, c = 0;
//		if(n % k == 0) need = k/2 + 1;
//		for(int i = 0; i < n; i++) {
//			if(vals[i] >= median && (i - c) % k == 0) c++;
//		}
//		if(c >= need) return true;
//		return false;
//	}

}

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

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

Think step by step.

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

1. **Edge Cases for `n` and `k`**:
   - When `n` is exactly `k` or a multiple of `k`.
   - When `n` is slightly more than `k` (e.g., `n = k + 1`).

2. **Edge Cases for Array Values**:
   - Arrays with all elements being the same.
   - Arrays with a mix of very high and very low values.
   - Arrays with a single large value and the rest being small values.

3. **Specific Test Cases**:
   - Test case where the median is the largest value in the array.
   - Test case where the median is the smallest value in the array.
   - Test case where the median is in the middle of the array.

Given these considerations, let's construct a small-scale test input that can expose the bug:

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

### Explanation of Test Cases:

1. **Test Case 1**:
   - `n = 3`, `k = 3`
   - Array: `[1, 2, 3]`
   - Since `n` is exactly `k`, the entire array will be removed in one operation, leaving an empty array. The median of an empty array is undefined, but the code should handle this case gracefully.

2. **Test Case 2**:
   - `n = 4`, `k = 2`
   - Array: `[1, 100, 100, 1]`
   - The possible final arrays after operations are `[1, 100]`, `[100, 1]`, and `[100, 100]`. The medians are `50.5`, `50.5`, and `100` respectively. The largest median is `100`.

3. **Test Case 3**:
   - `n = 5`, `k = 3`
   - Array: `[1, 2, 3, 4, 5]`
   - The possible final arrays after operations are `[1, 2]`, `[2, 3]`, `[3, 4]`, and `[4, 5]`. The medians are `1.5`, `2.5`, `3.5`, and `4.5` respectively. The largest median is `4.5`.

This test input should help expose any issues in the buggy code, particularly in handling edge cases and ensuring the correct median is calculated.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Med-imize
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given two positive integers $n$ and $k$, and another array $a$ of $n$ integers.

In one operation, you can select any subarray of size $k$ of $a$, then remove it from the array without changing the order of other elements. More formally, let $(l, r)$ be an operation on subarray $a_l, a_{l+1}, \ldots, a_r$ such that $r-l+1=k$, then performing this operation means replacing $a$ with $[a_1, \ldots, a_{l-1}, a_{r+1}, \ldots, a_n]$.

For example, if $a=[1,2,3,4,5]$ and we perform operation $(3,5)$ on this array, it will become $a=[1,2]$. Moreover, operation $(2, 4)$ results in $a=[1,5]$, and operation $(1,3)$ results in $a=[4,5]$.

You have to repeat the operation while the length of $a$ is greater than $k$ (which means $|a| \gt k$). What is the largest possible median$^\dagger$ of all remaining elements of the array $a$ after the process?

$^\dagger$The median of an array of length $n$ is the element whose index is $\left \lfloor (n+1)/2 \right \rfloor$ after we sort the elements in non-decreasing order. For example: $median([2,1,5,4,3]) = 3$, $median([5]) = 5$, and $median([6,8,2,4]) = 4$.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 5 \cdot 10^5$).

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

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

Output: For each test case, print a single integer — the largest median possible after performing the operations.

Examples:
input:
5
4 3
3 9 9 2
5 3
3 2 5 6 4
7 1
5 9 2 6 5 4 6
8 2
7 1 2 6 8 3 4 5
4 5
3 4 5 6
output:
3
4
9
6
4

Note: In the first test case, you can select a subarray $(l, r)$ which can be either $(1, 3)$ or $(2, 4)$. Thus, two obtainable final arrays are $[3]$ and $[2]$. The former one has the larger median ($3 > 2$) so the answer is $3$.

In the second test case, three obtainable final arrays are $[6, 4]$, $[3, 4]$, and $[3, 2]$. Their medians are $4$, $3$, and $2$ respectively. The answer is $4$.

In the third test case, only one element is left in the final array and it can be any element of the initial array. The largest one among them is $9$, so the answer is $9$.

Buggy Code:
import java.util.*;

public class main {
 public static int ans = 0;
 public static ArrayList<Integer> al = new ArrayList<>();

 public static int fun(int a[], int n, int i, int rem, int k) {
  if (rem == 0) {
   if (al.size() == 0) {
    return -1;
   }
   ArrayList<Integer> temp = new ArrayList<>();
   for (int x : al) {
    temp.add(x);
   }
   Collections.sort(temp);
   int medianidx = (temp.size() - 1) / 2;
   ans = Math.max(ans, temp.get(medianidx));
   return 0;
  }
  if (i >= n) {
   return -1;
  }
  for (int idx = i; idx < n; idx += k) {
   al.add(a[idx]);
   int rr = fun(a, n, idx + 1, rem - 1, k);
   if (rr == -1) {
    break;
   }
   al.remove(al.size() - 1);
  }
  return 0;
 }

 public static void main(String args[]) {
  Scanner sc = new Scanner(System.in);
  int t = sc.nextInt();
  while (t-- > 0) {
   int n = sc.nextInt(), k = sc.nextInt(), a[] = new int[n];
   int r = n % k;
   if (r == 0) {
    r = k;
   }
   for (int i = 0; i < n; i++) {
    a[i] = sc.nextInt();
   }
   ans = 0;
   fun(a, n, 0, r, k);
   System.out.println(ans);
  }
 }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
#include <bits/stdc++.h>
#define fastio {ios_base::sync_with_stdio(false); cin.tie(NULL); }
#define ll long long int
#define yes cout << "YES" << endl;
#define no cout << "NO" << endl;
#define sp " "
using namespace std;

int main() {
  fastio;
   int t;
   cin>>t;
   while(t--)
   {
    ll  n,k;
    cin>>n>>k;
    ll  sum=0;
    vector < ll > ar;
    for(ll i=0;i<k;i++)
    {
        ll c,p;
        cin>>c>>p;
        ar.push_back(p);
        sum+=p;

    }
   if(n>=k) cout<<sum<<endl;
   else
   {
       for(ll i=0;i<k-n;i++)
       {
           sum-=ar[i];
       }
       cout<<sum<<endl;
   }

   }
   return 0;

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The buggy code incorrectly assumes that the maximum sum can be achieved by simply removing the smallest `k - n` elements from the sum of all elements. This approach does not consider the constraint that all bottles on the same shelf must be of the same brand.

### Step-by-Step Analysis:
1. **Edge Case with Different Brands:**
   - If all bottles are of different brands, the code might incorrectly remove some bottles, even though they should be placed on different shelves.

2. **Edge Case with Fewer Shelves:**
   - If the number of shelves is less than the number of unique brands, the code might remove bottles that should be placed on separate shelves.

3. **Edge Case with High Cost Bottles:**
   - If there are high-cost bottles, the code might incorrectly remove them, leading to a suboptimal solution.

### Challenging Test Input:
Let's create a test input with a small number of test cases that includes the above scenarios:

1. **Test Case 1:**
   - 3 shelves, 4 bottles with different brands and different costs.
   - This will test the scenario where the code might remove bottles incorrectly.

2. **Test Case 2:**
   - 2 shelves, 3 bottles with the same brand but different costs.
   - This will test the scenario where the code might remove the highest cost bottle, which should not happen.

3. **Test Case 3:**
   - 1 shelf, 2 bottles with the same brand but different costs.
   - This will test the scenario where the code might remove the highest cost bottle, which should not happen.

### Test Input:
```plaintext
3
3 4
1 10
2 20
3 30
4 40
2 3
1 100
1 50
1 25
1 2
1 1000
1 500
```

### Explanation:
- **Test Case 1:**
  - 3 shelves, 4 bottles.
  - Brands: 1, 2, 3, 4.
  - Costs: 10, 20, 30, 40.
  - The correct answer is 100 (sum of all costs) because each bottle can be placed on a separate shelf.

- **Test Case 2:**
  - 2 shelves, 3 bottles.
  - Brands: 1, 1, 1.
  - Costs: 100, 50, 25.
  - The correct answer is 150 (sum of the highest two costs) because the two highest cost bottles can be placed on separate shelves.

- **Test Case 3:**
  - 1 shelf, 2 bottles.
  - Brands: 1, 1.
  - Costs: 1000, 500.
  - The correct answer is 1500 (sum of both costs) because both bottles can be placed on the same shelf.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

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

int  main()
{
    ios_base::sync_with_stdio(false), cin.tie(0);
    int t;
    cin >> t;
    while(t--){

        int n , k ;
    cin >> n >> k ;
    vector<int>v ;
    for(int i = 0 ; i < k ; i++){
        int x , y ;
        cin >> x >> y ;
        v.push_back(y) ;
    }

    sort(v.rbegin(),v.rend()) ;

    int sum = 0 ;
    if(n<k){
    for(int i=0 ; i<n ;i++){
        sum+=v[i] ;
    }
    }
    else
    {
         for(int i=0 ; i<k ;i++){
          sum+=v[i] ;
    }

    }

    cout << sum << endl ;

    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
1 3
1 10
2 20
3 30
2 3
1 10
2 20
3 30
3 3
1 10
2 20
3 30
10 1
1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
#define tc int t; cin>>t; while(t--)
#define fr(a,b)  for(int i=a;i<b;i++)
#define sfr(a,b)  for(int j=a;j<b;j++)
#define rfr(b,a) for(int i=b;i>=a;i--)
#define pb push_back
#define ll long long
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include<climits>
using namespace std;

#define ll long long
#define pb push_back
#define mp make_pair
#define pii pair<int, int>
#define vi vector<int>
#define vl vector<ll>
#define all(x) x.begin(), x.end()

const int INF = 1e9;
const ll LLINF = 1e18;
const int MOD = 1e9 + 7;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin>>t;
    while(t--){
    int n;
    cin>>n;
    int k;
    cin>>k;

    int ind[k];
    int pri[k];
    int in=0;
    int con = k;
    while(con--){
        cin>>ind[in];
        cin>>pri[in];
        in++;
    }

    int sum = 0;

    if(n >= k){
        for(int i=0;i<k;i++){
            sum += pri[i];
        }
    }
    else{
        sort(pri,pri+k,greater<int>());
        for(int i=0;i<n;i++){
            sum += pri[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 expose the bug in the provided code, we need to focus on scenarios where the code's logic might fail. The code sorts the prices of the bottles and sums the top `n` prices if `n` is less than `k`. However, it does not consider the constraint that all bottles on the same shelf must be of the same brand.

Let's construct a test case where this constraint is crucial:

1. **Test Case Description:**
   - We will have a small number of shelves and bottles.
   - We will have multiple bottles of different brands with different costs.
   - The optimal solution should place bottles of the same brand on the shelves to maximize the total cost.

2. **Test Case Input:**
   - Number of test cases: 1
   - Number of shelves: 3
   - Number of bottles: 5
   - Bottles:
     - Brand 1, Cost 10
     - Brand 1, Cost 20
     - Brand 2, Cost 30
     - Brand 2, Cost 40
     - Brand 3, Cost 50

3. **Expected Output:**
   - The maximum amount that can be earned is 90. This is achieved by placing:
     - Brand 1 (10 + 20) on one shelf (total 30)
     - Brand 2 (30 + 40) on another shelf (total 70)
     - Brand 3 (50) on the third shelf (total 50)

4. **Test Case Input:**
```plaintext
1
3 5
1 10
1 20
2 30
2 40
3 50
```

This test case should expose the bug because the code will sort the prices and sum the top 3, which will be 50, 40, and 30, leading to a total of 120. However, the correct answer should be 90, as the bottles must be placed such that all bottles on the same shelf are of the same brand.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
# import sys
# import os
# import math
# from collections import defaultdict

# if os.path.exists('B_input.txt'):
#     sys.stdin = open('B_input.txt', 'r')

# data = iter(sys.stdin.read().strip().splitlines())

# def input():
#     return next(data)

# while True:
#     try:
#         t = int(input())
#         for _ in range(t):
#             n, k = map(int, input().split())
#             val = defaultdict(int)
#             for _1 in range(k):
#                 b, c = map(int, input().split())
#                 val[b] += c
#             val = dict(sorted(val.items(), key=lambda item: item[1], reverse=True))
#             i = 0
#             ans = 0
#             for k, v in val.items():
#                 ans += v
#                 n-=1
#                 if n == 0:
#                     break

#             print(ans)
#     except Exception as e:
#         print(e)
#         break

import sys
import os
import math
from collections import defaultdict

if os.path.exists('B_input.txt'):
    sys.stdin = open('B_input.txt', 'r')

data = iter(sys.stdin.read().strip().splitlines())

def input():
    return next(data)

while True:
    try:
        t = int(input())
        for _ in range(t):
            n, k = map(int, input().split())
            val = [0]*(200005)
            for _1 in range(k):
                b, c = map(int, input().split())
                val[b] += c
            fval = []
            for v in val:
                if v:
                    fval.append(v)
            fval.sort(reverse=True)
            # print(val)
            i = 0
            ans = 0
            while n and i < len(fval):
                ans += fval[i]
                n-=1
                i+=1
            print(ans)
    except Exception as e:
        print(e)
        break

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that might cause issues. Here are some scenarios to consider:

1. **Maximum Number of Shelves and Bottles**: This can test the limits of the input constraints.
2. **Multiple Bottles of the Same Brand with Different Costs**: This can test how the code handles aggregation and sorting.
3. **Shelves Greater than the Number of Unique Brands**: This can test how the code handles unused shelves.
4. **All Bottles of the Same Brand**: This can test how the code handles a single brand.
5. **Shelves Less than the Number of Unique Brands**: This can test how the code handles the selection of the top brands.

Given the constraints and the need for a small-scale test input, let's create a test case that combines some of these scenarios:

```plaintext
5
1 2
1 10
1 20
2 3
1 10
2 20
3 30
3 4
1 10
2 20
3 30
4 40
4 5
1 10
2 20
3 30
4 40
5 50
1 1
1 1000
```

### Explanation:
1. **Test Case 1**:
   - 1 shelf, 2 bottles of the same brand with different costs.
   - Expected output: 30 (the sum of both bottles).

2. **Test Case 2**:
   - 2 shelves, 3 bottles of different brands.
   - Expected output: 50 (the sum of the two most expensive bottles).

3. **Test Case 3**:
   - 3 shelves, 4 bottles of different brands.
   - Expected output: 90 (the sum of the three most expensive bottles).

4. **Test Case 4**:
   - 4 shelves, 5 bottles of different brands.
   - Expected output: 140 (the sum of the four most expensive bottles).

5. **Test Case 5**:
   - 1 shelf, 1 bottle.
   - Expected output: 1000 (the cost of the single bottle).

This test input should help expose any issues in the provided code, particularly in how it handles multiple bottles of the same brand, different numbers of shelves, and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
t = int(input())  # количество наборов данных

for _ in range(t):
    n, k = map(int, input().split())  # количество полок и бутылок
    brand_prices = {}

    # Читаем данные о бутылках
    for _ in range(k):
        b, c = map(int, input().split())
        if b in brand_prices:
            brand_prices[b].append(c)
        else:
            brand_prices[b] = [c]

    # Считаем максимальную прибыль
    all_prices = []
    for prices in brand_prices.values():
        prices.sort(reverse=True)
        for price in prices:
            all_prices.append(price)

    # Сортируем все стоимости и выбираем самые дорогие
    all_prices.sort(reverse=True)
    max_profit = sum(all_prices[i] for i in range(min(n, len(all_prices))))

    print(max_profit)

**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
2 5
1 10
1 20
2 30
2 40
3 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
for _ in range(int(input())):
    n, k = map(int, input().split())
    brd = {}

    for _ in range(k):
        b, c = map(int, input().split())
        if b in brd:
            brd[b].append(c)
        else:
            brd[b] = [c]

    ttl = 0
    for i in brd.values():
        ttl += sum(i)

    if len(brd) <= n:
        print(ttl)
    else:
        maxi = []
        for i in brd.values():
            maxi.append(max(i))
        maxi.sort(reverse=True)
        maxin = sum(maxi[:n])
        print(maxin)

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

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

Think step by step.

Response:
```plaintext
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
import java.util.*;

public class VendingMachineProfit {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();  // Number of test cases

        while (t-- > 0) {
            int n = scanner.nextInt();  // Number of shelves
            int k = scanner.nextInt();  // Number of bottles

            // Map to store bottles of each brand and their costs
            Map<Integer, List<Integer>> brandCosts = new HashMap<>();

            for (int i = 0; i < k; i++) {
                int brand = scanner.nextInt();
                int cost = scanner.nextInt();

                brandCosts.computeIfAbsent(brand, x -> new ArrayList<>()).add(cost);
            }

            // List to store the maximum profit we can get from each brand
            List<Integer> maxProfits = new ArrayList<>();

            for (List<Integer> costs : brandCosts.values()) {
                // Sort costs of each brand in descending order
                costs.sort(Collections.reverseOrder());

                // Calculate the profit of taking the top bottles of this brand
                int profit = 0;
                for (int i = 0; i < Math.min(n, costs.size()); i++) {
                    profit += costs.get(i);
                }
                maxProfits.add(profit);
            }

            // Sort the max profits in descending order and take the top n profits
            maxProfits.sort(Collections.reverseOrder());

            int totalProfit = 0;
            for (int i = 0; i < Math.min(n, maxProfits.size()); i++) {
                totalProfit += maxProfits.get(i);
            }

            System.out.println(totalProfit);
        }

        scanner.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 6
1 10
1 20
2 30
2 40
3 50
3 60
```

### Explanation:
This test case is designed to expose a potential flaw in the logic of the provided buggy code. Here's why:

1. **Multiple Brands with Multiple Bottles:** The test case includes multiple brands (1, 2, 3) each with multiple bottles. This ensures that the code must correctly handle the aggregation of costs for each brand and then select the best combination of brands to maximize the total profit.

2. **Different Cost Distributions:** The costs for each brand are different and vary significantly. For example, brand 1 has costs 10 and 20, brand 2 has costs 30 and 40, and brand 3 has costs 50 and 60. This setup ensures that the code must correctly sort and sum the costs for each brand.

3. **Number of Shelves and Bottles:** There are 3 shelves and 6 bottles. This means that the code must correctly handle the scenario where the number of shelves is less than the number of bottles, and it must select the best combination of bottles to place on the shelves.

4. **Expected Output:** The expected output for this test case is 130. The optimal solution is to place the highest cost bottles from each brand on the shelves:
   - Brand 1: 20
   - Brand 2: 40
   - Brand 3: 60
   The total profit is 20 + 40 + 60 = 130.

This test case is designed to ensure that the code correctly handles the sorting and selection of the highest cost bottles for each brand and then selects the best combination to maximize the total profit.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
import java.util.*;

public class demo {

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

        while (t-- > 0) {
            int n = scanner.nextInt();
            int k = scanner.nextInt();

            Map<Integer, List<Integer>> brandCosts = new HashMap<>();

            for (int i = 0; i < k; i++) {
                int bi = scanner.nextInt();
                int ci = scanner.nextInt();

                brandCosts.putIfAbsent(bi, new ArrayList<>());
                brandCosts.get(bi).add(ci);
            }

            List<Integer> maxCosts = new ArrayList<>();

            for (List<Integer> costs : brandCosts.values()) {
                Collections.sort(costs, Collections.reverseOrder());
                for (int cost : costs) {
                    maxCosts.add(cost);
                }
            }

            Collections.sort(maxCosts, Collections.reverseOrder());
            int totalEarnings = 0;
            for (int i = 0; i < Math.min(n, maxCosts.size()); i++) {
                totalEarnings += maxCosts.get(i);
            }

            System.out.println(totalEarnings);
        }

    }
}

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

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

Think step by step.

Response:
```plaintext
1
1 2
1 1000
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
#include <iostream>
#include <vector>
#include <unordered_map>
#include <queue>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;

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

        unordered_map<int, priority_queue<int>> brand_costs;
        for (int i = 0; i < k; ++i) {
            int b, c;
            cin >> b >> c;
            brand_costs[b].push(c);
        }

        priority_queue<int> max_heap;

        // Fill the max heap with the maximum cost from each brand
        for (auto &pair : brand_costs) {
            if (!pair.second.empty()) {
                max_heap.push(pair.second.top());
            }
        }

        int max_amount = 0;

        // Extract maximum costs up to n times (for n shelves)
        for (int i = 0; i < n && !max_heap.empty(); ++i) {
            int current_max = max_heap.top();
            max_heap.pop();
            max_amount += current_max;

            // If there's another bottle from the same brand, push it to the heap
            for (auto &pair : brand_costs) {
                if (!pair.second.empty() && pair.second.top() == current_max) {
                    pair.second.pop(); // Remove the current max cost
                    if (!pair.second.empty()) {
                        max_heap.push(pair.second.top()); // Push the next max cost
                    }
                    break; // Exit the loop after handling the current max
                }
            }
        }

        cout << max_amount << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
3 5
1 10
1 10
2 5
2 5
3 1
1 1000
5 5
1 10
2 20
3 30
4 40
5 50
10 1
1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int, int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vll vector<long long>
#define vvll vector<vector<long long>>
#define mii map<int, int>
#define msi map<string, int>
#define mci map<char, int>
#define si set<int>
#define sc set<char>
#define f(i, s, e) for (long long int i = s; i < e; i++)
#define pb push_back
#define mod 1000000007

void solve()
{
    ll n,k;
    cin>>n>>k;
    vvll v(k,vll(2));
    priority_queue<ll> pq;
    f(i,0,k){
        cin>>v[i][0]>>v[i][1];
        pq.push(v[i][1]);
    }
    ll sum=0;
    while(!pq.empty() && n>0){
        sum+=pq.top();
        pq.pop();
        n--;
    }
    cout<<sum<<endl;

}
int main()
{

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 4
1 10
1 20
2 15
2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
def max_earnings(t, test_cases):
    results = []

    for case in test_cases:
        n, k, bottles = case
        brand_costs = {}

        # Collect costs for each brand
        for bi, ci in bottles:
            if bi not in brand_costs:
                brand_costs[bi] = []
            brand_costs[bi].append(ci)

        # Gather the highest possible earnings per brand
        max_brand_costs = []
        for costs in brand_costs.values():
            costs.sort(reverse=True)
            max_brand_costs.append(sum(costs[:min(n, len(costs))]))

        # Sort the earnings by value and take the top n shelves
        max_brand_costs.sort(reverse=True)
        max_earn = sum(max_brand_costs[:n])

        results.append(max_earn)

    return results

def solve():
    t = int(input())
    test_cases = []

    for _ in range(t):
        n, k = map(int, input().split())
        bottles = [tuple(map(int, input().split())) for _ in range(k)]
        test_cases.append((n, k, bottles))

    results = max_earnings(t, test_cases)

    for result in results:
        print(result)

solve()

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases that can highlight issues in the logic. The buggy code has a few potential issues:

1. **Handling of multiple bottles of the same brand on different shelves**: The code should ensure that the maximum cost for each brand is considered, and the top `n` values are summed correctly.
2. **Edge cases with a large number of shelves and a small number of bottles**: The code should handle scenarios where the number of shelves is much larger than the number of bottles.
3. **Edge cases with a large number of bottles and a small number of shelves**: The code should handle scenarios where the number of bottles is much larger than the number of shelves.
4. **Handling of duplicate brands**: The code should correctly aggregate and sort the costs for each brand.

Let's create a test input that covers these edge cases:

1. **Test case with multiple bottles of the same brand on different shelves**:
   - This will test if the code correctly aggregates the costs for each brand and selects the top `n` values.

2. **Test case with a large number of shelves and a small number of bottles**:
   - This will test if the code handles the scenario where there are more shelves than bottles.

3. **Test case with a large number of bottles and a small number of shelves**:
   - This will test if the code correctly aggregates the costs and selects the top `n` values.

4. **Test case with duplicate brands**:
   - This will test if the code correctly aggregates the costs for each brand and handles duplicates.

Here is the test input:

```plaintext
4
3 3
2 6
2 7
1 15
190000 1
1 1000
1 3
2 6
2 7
1 15
6 2
1 7
2 5
```

Explanation:
1. **Test case 1**:
   - 3 shelves, 3 bottles.
   - Bottles: (2, 6), (2, 7), (1, 15).
   - Expected output: 28 (2 bottles of brand 2 on one shelf and 1 bottle of brand 1 on another shelf).

2. **Test case 2**:
   - 190000 shelves, 1 bottle.
   - Bottle: (1, 1000).
   - Expected output: 1000 (only one bottle, so it goes on one shelf).

3. **Test case 3**:
   - 1 shelf, 3 bottles.
   - Bottles: (2, 6), (2, 7), (1, 15).
   - Expected output: 15 (only one shelf, so the highest cost bottle is selected).

4. **Test case 4**:
   - 6 shelves, 2 bottles.
   - Bottles: (1, 7), (2, 5).
   - Expected output: 12 (both bottles are placed on separate shelves).

This test input covers various edge cases and should help expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
t = int(input())
results = []

for _ in range(t):
    n, k = map(int, input().split())
    prices = []

    for _ in range(k):
        b, c = map(int, input().split())
        prices.append(c)
    prices.sort(reverse=True)

    max_sum = sum(prices[:n])
    results.append(max_sum)

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 expose the bug in the provided buggy code, we need to focus on the logic that the code uses to determine the maximum amount Arseniy can earn. The buggy code sorts all the bottle costs in descending order and then sums the first `n` costs, assuming that the highest costs will always be the best to place on the shelves. However, this approach ignores the constraint that all bottles on the same shelf must be of the same brand.

### Step-by-Step Analysis:
1. **Identical Brand Constraint**: The code does not consider that all bottles on the same shelf must be of the same brand. This means that simply taking the top `n` costs without regard to brand will often lead to an incorrect solution.

2. **Edge Cases**:
   - **Single Brand**: If all bottles are of the same brand, the code should work correctly.
   - **Multiple Brands**: If there are multiple brands, the code will fail if the top `n` costs come from different brands.

3. **Test Input Design**:
   - **Small Scale**: We need a small-scale test input to ensure it fits within the constraints and is easy to verify.
   - **Multiple Brands**: We should include a test case where the top `n` costs are from different brands to expose the bug.

### Test Input:
Let's design a test case where the top `n` costs are from different brands, but the optimal solution involves placing the highest-cost bottles of the same brand on the shelves.

```plaintext
1
4 6
1 10
2 20
3 30
4 40
1 50
2 60
```

### Explanation:
- **Test Case**:
  - `n = 4` (4 shelves)
  - `k = 6` (6 bottles)
- **Bottles**:
  - Brand 1: 10, 50
  - Brand 2: 20, 60
  - Brand 3: 30
  - Brand 4: 40

### Expected Output:
- **Correct Solution**:
  - The optimal solution is to place the two bottles of Brand 2 (60, 20) on two shelves and the two bottles of Brand 1 (50, 10) on the other two shelves.
  - Total cost: 60 + 20 + 50 + 10 = 140

- **Buggy Code Output**:
  - The buggy code will sort the costs in descending order: 60, 50, 40, 30
  - It will sum the top 4 costs: 60 + 50 + 40 + 30 = 180
  - This is incorrect because it places bottles of different brands on the same shelves.

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: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
import java.util.*;

public class Startup {
    public static List<Integer> maxEarnings(int t, List<int[]> testCases, List<List<int[]>> bottleData) {
        List<Integer> results = new ArrayList<>();

        for (int i = 0; i < t; i++) {
            int n = testCases.get(i)[0];
            List<int[]> bottles = bottleData.get(i);
            Map<Integer, List<Integer>> brandCosts = new HashMap<>();

            for (int[] bottle : bottles) {
                int brand = bottle[0];
                int cost = bottle[1];
                brandCosts.putIfAbsent(brand, new ArrayList<>());
                brandCosts.get(brand).add(cost);
            }

            List<Integer> maxBrandEarnings = new ArrayList<>();
            for (List<Integer> costs : brandCosts.values()) {
                costs.sort(Collections.reverseOrder());
                int total = 0;
                for (int j = 0; j < Math.min(n, costs.size()); j++) {
                    total += costs.get(j);
                }
                maxBrandEarnings.add(total);
            }

            maxBrandEarnings.sort(Collections.reverseOrder());
            int maxEarning = 0;
            for (int j = 0; j < Math.min(n, maxBrandEarnings.size()); j++) {
                maxEarning += maxBrandEarnings.get(j);
            }
            results.add(maxEarning);
        }

        return results;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        List<int[]> testCases = new ArrayList<>();
        List<List<int[]>> bottleData = new ArrayList<>();

        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            testCases.add(new int[]{n, k});

            List<int[]> bottles = new ArrayList<>();
            for (int j = 0; j < k; j++) {
                int brand = sc.nextInt();
                int cost = sc.nextInt();
                bottles.add(new int[]{brand, cost});
            }
            bottleData.add(bottles);
        }

        List<Integer> results = maxEarnings(t, testCases, bottleData);
        for (int result : results) {
            System.out.println(result);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
```

This test input is designed to cover a variety of scenarios that can expose potential issues in the buggy code:

1. **Test Case 1:**
   - **Input:** 3 shelves, 3 bottles.
   - **Bottles:** Brand 2 (6), Brand 2 (7), Brand 1 (15).
   - **Expected Output:** 28.
   - **Reason:** This tests the ability to place the most expensive bottles on different shelves.

2. **Test Case 2:**
   - **Input:** 1 shelf, 3 bottles.
   - **Bottles:** Brand 2 (6), Brand 2 (7), Brand 1 (15).
   - **Expected Output:** 15.
   - **Reason:** This tests the scenario where only one shelf is available, and the most expensive bottle should be chosen.

3. **Test Case 3:**
   - **Input:** 6 shelves, 2 bottles.
   - **Bottles:** Brand 1 (7), Brand 2 (5).
   - **Expected Output:** 12.
   - **Reason:** This tests the scenario where the number of shelves is greater than the number of bottles, ensuring that all bottles are placed.

4. **Test Case 4:**
   - **Input:** 190000 shelves, 1 bottle.
   - **Bottles:** Brand 1 (1000).
   - **Expected Output:** 1000.
   - **Reason:** This tests the scenario with a very large number of shelves and only one bottle, ensuring the code handles large inputs correctly.

These test cases are designed to ensure that the code can handle different numbers of shelves and bottles, and that it correctly maximizes the earnings based on the constraints provided.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
import java.util.*;
public class n {
    public static void main(String[] args) {
       Scanner sc=new Scanner(System.in);
       int t=sc.nextInt();
       for(int i=0;i<t;i++){
        int n=sc.nextInt();
        int k=sc.nextInt();
        int[]a=new int[k];
        int[]b=new int[k];
        ArrayList<Integer> c=new ArrayList<Integer>();
        for(int j=0;j<k;j++){
            a[j]=sc.nextInt();
            b[j]=sc.nextInt();
        }
       // Arrays.sort(a);
       for(int j=0;j<k;j++){
            int count=0;
            for(int f=j;f>=0;f--){
               if(a[j]==a[f] && j!=f){
                count++;}
               }
               if(count==0){
               int sum=b[j];
            for(int l=j+1;l<k;l++){
                if(a[j]==a[l]){
                count++;
                sum=sum+b[l];
                }
            }
            count++;
            c.add(sum);
        }
        }
        Collections.sort(c);
        int s=0;
        int size=c.size();
        //int z=k-size;
        if(n==k){
            for(int w=0;w<size;w++){
                s=s+c.get(size-w-1);
            }
            System.out.println(s);
           }
        else if(n<k){
        for(int w=0;w<n;w++){
            s=s+c.get(size-w-1);
        }
        System.out.println(s);
       }
       else if(n>k){
        for(int w=0;w<size;w++){
            s=s+c.get(size-w-1);
        }
        System.out.println(s);
    }
       }
    }
 }

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

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

Think step by step.

Response:
```plaintext
4
1 3
1 1000
1 1000
1 1000
3 3
1 1000
2 1000
3 1000
2 4
1 1000
1 1000
2 1000
2 1000
1 1
1 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
t = int(input())

for z in range(t):
    n, k = map(int, input().split())
    soda = {}
    ai = [0] * k
    for i in range(k):
        bi, ci = map(int, input().split())

        ai[bi - 1] += ci

    print(sum(ai[:n]))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code might fail. The primary issue with the buggy code is that it incorrectly accumulates the costs of bottles by brand and then sums the first `n` elements of the list, which doesn't necessarily give the maximum possible earnings.

### Key Points to Consider:
1. **Multiple Brands with Different Costs**: The code should handle multiple brands with different costs and ensure that the highest-cost brands are placed on the shelves.
2. **Shelves More than Bottles**: If the number of shelves is greater than the number of different brands, the code should handle this scenario correctly.
3. **Single Brand with Multiple Bottles**: If there are multiple bottles of the same brand, the code should sum their costs correctly.
4. **Edge Cases**: Test with the minimum and maximum values for `n` and `k`.

### Challenging Test Input:
Let's create a test case where the number of shelves is greater than the number of different brands, and the brands have different costs. This will highlight the issue with the current approach of summing the first `n` elements.

```plaintext
3
3 2
1 10
2 5
1 2
1 10
2 5
6 3
1 10
2 5
3 1
```

### Explanation:
1. **First Test Case**:
   - 3 shelves, 2 bottles.
   - Brands: 1 (cost 10), 2 (cost 5).
   - The optimal placement is to place the bottle of brand 1 on one shelf and the bottle of brand 2 on another shelf. The total cost is 10 + 5 = 15. The code should output 15.

2. **Second Test Case**:
   - 1 shelf, 2 bottles.
   - Brands: 1 (cost 10), 2 (cost 5).
   - The optimal placement is to place the bottle of brand 1 on the shelf. The total cost is 10. The code should output 10.

3. **Third Test Case**:
   - 6 shelves, 3 bottles.
   - Brands: 1 (cost 10), 2 (cost 5), 3 (cost 1).
   - The optimal placement is to place the bottle of brand 1 on one shelf, the bottle of brand 2 on another shelf, and the bottle of brand 3 on another shelf. The total cost is 10 + 5 + 1 = 16. The code should output 16.

This test input will help expose the bug in the buggy code by showing that the current approach of summing the first `n` elements of the list is incorrect and does not always yield the maximum possible earnings.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
from collections import defaultdict

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    brand_cost = defaultdict(list)

    for _ in range(k):
        b, c = map(int, input().split())
        brand_cost[b].append(c)

    whole_cost = []
    for costs in brand_cost.values():
        costs.sort(reverse=True)
        whole_cost.append(sum(costs[:n]))

    whole_cost.sort(reverse=True)
    total_earn = sum(whole_cost[:n])

    print(total_earn)

**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
2 4
1 10
1 20
2 30
2 40
1 4
1 10
1 20
1 30
1 40
3 3
1 10
2 20
3 30
```