**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
#include <iostream>
using namespace std;
int main() {
  int t,n,k,l=0;  cin>>t;
  for(int i=0;i<t;i++)
  {
      cin>>n>>k;   int c[n],fq[100]={0};
      for(int j=0;j<n;j++)
     {
         cin>>c[j];  fq[c[j]]++;
     }
        for(int j=0;j<100;j++)
     {
         if(fq[j]>=k) l++;
     }
     if(l==0) { cout<<n<<endl;  continue; }
     else cout<<k-1<<endl;
     l=0;
  }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
10 3
1 1 1 2 2 2 3 3 3 4
5 2
1 1 2 2 3
7 2
1 1 1 2 2 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
def minimize(valor, k):
    valor.sort()
    if valor[-1] < k:
        print(sum(valor))
        return
    else:
        total = 0
        for m in range(len(valor)):
            n = valor[m]//k
            total += n*(k-1)
            valor[m] = valor[m]%k
        valor.sort()
        for n in range(len(valor) - 1, -1,-1):
            if valor[n] + total <= k:
                valor[n] = valor[n] + total
                break
            else:
                total = total - (k - valor[n])
                valor[n] = k
                if n == 0:
                    valor[-1] = total
                    break
        minimize(valor, k)

for _ in range(int(input())):
    n_k = list(map(int, input().split()))
    n, k = n_k[0], n_k[1]
    cards = list(map(int, input().split()))
    contador = {}
    for num in cards:
        contador[num] = contador.get(num, 0) + 1
    valor = list(contador.values())
    minimize(valor, k)

**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
9 3
3 3 3 3 3 3 3 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
def cards(n,k,arr):
    arr.sort()
    record=[]
    cnt=1
    record = []
    cnt = 1

    for i in range(n-1):
        if arr[i] == arr[i+1]:
            cnt += 1
        else:
            record.append(cnt)
            cnt = 1

    # Append the count for the last sequence
    record.append(cnt)

    if max(record)>=k and n>k:
        print(k-1)
    else:
        print(n)

d = int(input())
for _ in range(d):
    nk = input().split()
    n = int(nk[0])
    k = int(nk[1])
    ar= input().split()
    arr= list(map(int, ar))
    cards(n,k,arr)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
import heapq
#from collections import defaultdict
from collections import Counter
#import math as m
#import numpy as np
#map(int,input().split())
#list(map(int,input().split()))
def solve():
    n,k = map(int,input().split())
    c = list(map(int,input().split()))
    cnt = Counter(c)
    lst = []
    for v in cnt.values():
        lst.append([k-v%k if v%k != 0 else 0,v])
    lst.sort()
    while n >= k:
        e = lst.pop(0)
        if e[1] < k:
            return n
        r = e[1]//k
        n-=r
        e[1] = e[1]%k
        e[0] = k - e[1]%k
        if e[1] != 0:
            heapq.heappush(lst,e)
        left = r*(k-1)
        if len(lst)==0:
            e[1] = left
            e[0] = k - e[1]%k if e[1]%k != 0 else 0
            lst.append(e)
        for i in range(len(lst)):
            if lst[i][0] == 0:
                continue
            lst[i][0] -= left
            if lst[i][0] < 0:
                lst[i][1] += left + lst[i][0]
                left = abs(lst[i][0])
                lst[i][0] = 0
            else:
                lst[i][1] += left
                left = 0
            if left >= k:
                r = left//k
                lst[i][1] += r*k
                left -= r*k
            if left == 0:
                break
        if left != 0:
            if len(lst) == 0:
                return left
            e = lst.pop()
            e[1] = e[1] + left
            e[0] = k - e[1]%k if e[1]%k != 0 else 0
            heapq.heappush(lst,e)
    return n

t=int(input())
for _ in range(t):
    print(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
10 2
1 1 1 1 1 1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
import java.util.*;
public class A443 {
    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[] arr= new int[n];
            for (int j = 0; j < arr.length; j++) {
                arr[j]=sc.nextInt();
            }
            if(n<k) {System.out.println(1); continue;}
            Arrays.sort(arr);
            HashMap<Integer, Integer> map=new HashMap<>();
            int a=arr[0];
            map.put(arr[0], 1);
            for (int j = 1; j < arr.length; j++) {
                if(arr[j]==a) map.put(a, map.get(a)+1);
                else{
                    a=arr[j];
                    map.put(arr[j], 1);
                }
            }int count=0,b=0,cnt=0;
            for (int j : map.values()) {
                if(j>=k) count++;
                else if(j==k-1) cnt++;
                else b++;
            }
            if(count==1 && b==0)
            System.out.println(k-1);
            else if(count==0)
            System.out.println(n);
            else if(b>=k)
            System.out.println(n-k*count-1);
            else if(cnt>0 && count>0)
            System.out.println(k-1);
            else
            System.out.println(n-k*count);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
/*

    ॐ भूर् भुवः स्वः।
    तत् सवितुर्वरेण्यं।
    भर्गो देवस्य धीमहि।
    धियो यो नः प्रचोदयात् ॥

  @Author  : ANKUSH
  @Country : INDIA

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

public class main1 {

    private static int n(long l) {
        throw new UnsupportedOperationException("Not supported yet.");
    }
// Codechef {
    static class pair {
        int m;
        int val1;
        int val2;

        public pair(int m,int val1,int val2) {
            this.m = m;
            this.val1=val1;
            this.val2=val2;
        }
        /*
         * public boolean equals(Object o) {
         * if (this == o)
         * return true;
         * if (o == null || getClass() != o.getClass())
         * return false;
         * pair pair = (pair) o;
         * return val1 == pair.val1 && val2 == pair.val2;
         * }
         *
         * @Override
         * public int hashCode() {
         * return 31 * Long.hashCode(val1) + Long.hashCode(val2);
         * }
         *
         * @Override
         * public String toString() {
         * return "pair{" + "x=" + val1 + ", y=" + val2 + '}';
         * }
         *
         */
    }

    static long MOD = (long) (1e9 + 7);
    static long dp[];
    static long ans;
    public static void main(String[] args) {

        int TTT = 0;
        try {
            FastReader in = new FastReader();
            FastWriter out = new FastWriter();
            // preFactorial();
          int testCases = in.nextInt();
           while (TTT++ != testCases) {
                // write code here\
               int n=in.nextInt();
               int k=in.nextInt();
               int a[]=new int[n];
               int f[]=new int[101];
               for(int i=0;i<n;i++){
                a[i]=in.nextInt();
                f[a[i]]++;
               }
               boolean flag=false;
               for(int i=0;i<100;i++){
                if(f[i]>=k){
                    flag=true;
                    break;
                }
               }
               if(flag){
                out.println(k-1);
               }
               else{
                out.println(n);
               }

               }
          out.close();
        } catch (Exception e) {
            System.out.println(e);
            return;
        }
    }
    public static long cal(long i){
        long sum=0;
        while(i>0){
            int dig=(int)i%10;
            sum+=dig;
            i/=10;
        }
        return sum;
    }

    /*-------------------------------------------  MATHS     MATHS     MATHS   ------------------------------------ */
    static boolean isSquare(long n) {
		long sqrt=(long) Math.sqrt(n);
		for (int i=(int) Math.max(1, sqrt-5); i<=sqrt+5; i++)
			if (i*i==n)
				return true;
		return false;
	}
    public static boolean isPrime(long number) {
        if (number <= 1) {
            return false;
        }
        if (number <= 3) {
            return true;
        }
        if (number % 2 == 0 || number % 3 == 0) {
            return false;
        }

        long sqrt = (long) Math.sqrt(number);
        for (long i = 5; i <= sqrt; i += 6) {
            if (number % i == 0 || number % (i + 2) == 0) {
                return false;
            }
        }
        return true;
    }

    public static long power(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1)
                ans *= a;
            a *= a;
            b >>= 1;
        }
        return ans;
    }

    public static long power_mod(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1) {
                ans *= a;
                ans %= MOD;
            }
            a *= a;
            a %= MOD;
            b >>= 1;
        }
        return ans;
    }

    static final int SIZE = 1005;
    private static long[] fact = new long[SIZE];

    public static long add_mod(long a, long b) {
        return (a + b) % MOD;
    }

    public static long sub_mod(long a, long b) {
        return ((a - b) % MOD + MOD) % MOD;
    }

    public static long multiply_mod(long a, long b) {
        return (a * b) % MOD;
    }

    public static long inv(long a) {
        return power_mod(a, MOD - 2);
    }

    public static long divide_mod(long a, long b) {
        return multiply_mod(a, inv(b));
    }

    public static long nCr(long n, long r) {
        if (n < r)
            return 0;
        return divide_mod(fact[(int) n], multiply_mod(fact[(int) r], fact[(int) (n - r)]));
    }

    public static long nPr(long n, long r) {
        if (n < r)
            return 0;
        return divide_mod(fact[(int) n], fact[(int) (n - r)]);
    }

    public static void preFactorial() {
        fact[0] = 1;
        for (int i = 1; i < SIZE; i++) {
            fact[i] = multiply_mod(i, fact[i - 1]);
        }
    }

    public static void Sieve_Of_Eratosthenes() {
        boolean[] isPrime = new boolean[SIZE + 1];
        Arrays.fill(isPrime, true);
        isPrime[0] = false;
        isPrime[1] = false;

        for (int i = 2; i * i <= SIZE; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= SIZE; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }

    public static long gcd(long x, long y) {
        if (x > y)
            return gcd(y, x);
        if (y == 0) {
            return x;
        }
        if (x == 0) {
            return y;
        }
        return gcd(y % x, x);
    }

    public static long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }

    /*----------------------------------------  ADVANCED DATA STRUCTURES  ADVANCED DATA STRUCTURES  ------------------------------------ */
    public static class DisjointSet {
        private long[] parent;
        private long[] size;

        public DisjointSet(int n) {
            parent = new long[n];
            size = new long[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }

        public long findParent(long node) {
            if (node == parent[(int) node]) {
                return node;
            }
            return parent[(int) node] = findParent(parent[(int) node]);
        }

        public void unionSize(long node1, long node2) {
            long p1 = findParent(node1);
            long p2 = findParent(node2);
            if (p1 == p2)
                return;
            if (size[(int) p1] > size[(int) p2]) {
                size[(int) p1] += size[(int) p2];
                parent[(int) p2] = p1;
            } else {
                size[(int) p2] += size[(int) p1];
                parent[(int) p1] = p2;
            }
        }

        public long findSize(long node) {
            long p1 = findParent(node);
            return size[(int) p1];
        }
    }

    static class Segment_Tree {
        long arr[];
        int n;
        long seg[];

        public Segment_Tree(long arr[], int n) {
            this.arr = arr;
            this.n = n;
            this.seg = new long[(4 * n) + 1];
            build(0, 0, n - 1);
        }

        public void build(int ind, int low, int high) {
            if (low == high) {
                seg[ind] = arr[low];
                return;
            }
            int mid = (low + high) / 2;
            build((2 * ind) + 1, low, mid);
            build((2 * ind) + 2, mid + 1, high);
            seg[ind] = (seg[2 * ind + 1] ^ seg[2 * ind + 2]);
        }

        public long query2(int ind, int low, int high, int left, int right) {
            if (low >= left && high <= right) {
                return seg[ind];
            }
            if (low > right || high < left) {
                return 0;
            }
            int mid = (low + high) / 2;
            long l = query2(2 * ind + 1, low, mid, left, right);
            long r = query2(2 * ind + 2, mid + 1, high, left, right);
            return (l ^ r);
        }

        public long query(int l, int r) {
            return query2(0, 0, n - 1, l, r);
        }

    }

    /*-------------------------------------  FAST READER AND WRITER      FAST READER AND WRITER  ------------------------------------ */
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static class FastWriter {

        private final BufferedWriter bw;

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

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

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

}

/*-------------------------------------XXXXXXXX      THE END         THE END        XXXXXXXX------------------------------------ */

**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
10 3
1 1 1 1 1 1 1 1 1 1
6 3
1 2 3 4 5 6
7 2
1 1 1 1 1 1 1
```

### Explanation:

1. **Test Case 1:**
   - **Input:** `10 3` (10 cards, `k=3`), with all cards having the same number.
   - **Expected Output:** `2`
   - **Reason:** Since all cards are the same, we can perform the operation multiple times to reduce the number of cards. Each operation reduces the count by 1, so we can reduce 10 cards to 2 cards.

2. **Test Case 2:**
   - **Input:** `6 3` (6 cards, `k=3`), with all cards having different numbers.
   - **Expected Output:** `6`
   - **Reason:** Since no three cards have the same number, no operations can be performed, so the number of cards remains 6.

3. **Test Case 3:**
   - **Input:** `7 2` (7 cards, `k=2`), with all cards having the same number.
   - **Expected Output:** `1`
   - **Reason:** Since all cards are the same and `k=2`, we can perform the operation multiple times to reduce the number of cards. Each operation reduces the count by 1, so we can reduce 7 cards to 1 card.

These test cases cover different scenarios:
- A case where all cards are the same and `k` is greater than 1.
- A case where all cards are different.
- A case where all cards are the same and `k` is 2.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

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

#define ll             long long int
#define ulli           unsigned long long int
#define li             long int
#define fa(i,a,n)      for(int i=a;i<n;i++)
#define fb(i,n,a)      for(int i=n;i>=a;i--)
#define w(t)           while(t--)
#define l(s)           s.length()
#define fast           ios_base::sync_with_stdio(false);
#define sa(a,n)        sort(a,a+n);
#define sv(v)          sort(v.begin(),v.end())
#define cy             cout<<"YES"<<endl;
#define cn             cout<<"NO"<<endl;
#define nl             cout<<endl;
#define cmi            cout<<"-1"<<endl;
#define vi             vector<int>
#define pb             push_back
#define tc             int t; cin>>t;
#define pp             pair<int,int>
#define input(a,n)     for(int i=0;i<n;i++) cin>>a[i];
#define mod            1000000007
#define co(n)          cout<<n;
#define ret            return 0
#define mi             map<int,int>
#define output(a,n)    for(int i=0;i<n;i++) cout<<a[i]<<" ";
#define forn(i, n)     fa(i,0,n)
#define sz(v)          int((v).size())

using namespace std;

void asdf()
{
  int n,k;
  cin>>n>>k;
  int a[n+5];
  input(a,n)
  sort(a,a+n);
  int c=1;
  int m=1;
  for(int i=0;i<n-1;i++)
  {
      if(a[i]==a[i+1])
      {
          c++;
          if(c>m)
          {
              m=c;
          }
      }
      else
      {
          c=1;
      }
  }
  if(n<k)
  {
      cout<<"1"<<endl;
  }
  else if(k<=m)
  {
      cout<<k-1<<endl;
  }
  else
  {
      cout<<n<<endl;
  }
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        asdf();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

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

int main (){
    int q,n,k;
    cin >>q;
    for (int z=0;z<q;z++){
        cin>>n>>k;
        vector<int> arr(102, 0);
        int x;
        int flag=false;
        int flag2=0;
        int ans=0;
        for(int i=0;i<n;i++){
            cin>>x;
            arr[x]++;
        }
        for(int i=1;i<101;i++){
            if(arr[i]>=k)flag=true;
        }
        for(int i=1;i<101;i++){
            if(flag==true){
                if(arr[i]>=(k-1)){
                    if(flag2==0)
                    ans=ans+(k-1);
                    flag2=1;
                }
                else ans=ans+arr[i];
            }
            else {
                ans=ans+arr[i];
            }
        }
        cout<<ans<<endl;

    }
return 0;

}

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

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

Think step by step.

Response:
```plaintext
3
10 3
1 1 1 1 1 1 1 1 1 1
5 2
1 1 2 2 3
7 2
1 1 1 1 1 1 1
```

### Explanation:
1. **Test Case 1:**
   - **Input:** `10 3 1 1 1 1 1 1 1 1 1 1`
   - **Expected Output:** `1`
   - **Explanation:** You have 10 cards all with the number 1. You can perform the exchange operation 3 times, each time exchanging 3 cards for 2 new cards, until you are left with 1 card.

2. **Test Case 2:**
   - **Input:** `5 2 1 1 2 2 3`
   - **Expected Output:** `3`
   - **Explanation:** You have 5 cards with numbers 1, 1, 2, 2, 3. You can perform the exchange operation twice (once for the two 1s and once for the two 2s), leaving you with 3 cards (one 1, one 2, and one 3).

3. **Test Case 3:**
   - **Input:** `7 2 1 1 1 1 1 1 1`
   - **Expected Output:** `1`
   - **Explanation:** You have 7 cards all with the number 1. You can perform the exchange operation 3 times, each time exchanging 2 cards for 1 new card, until you are left with 1 card.

These test cases are designed to expose the bug in the provided buggy code by focusing on scenarios where the number of cards and the exchange size `k` are such that the algorithm should be able to reduce the number of cards significantly. The buggy code incorrectly handles the logic for reducing the number of cards, and these test cases will help highlight that issue.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    input_string = input()
    numbers = input_string.split()  # Разделить строку по пробелам
    numbers = [int(num) for num in numbers]  # Преобразовать каждое значение в целое число
    # print(numbers)
    dick = {}
    for i in numbers:
        dick[i] = numbers.count(i)

    sorted_dict = dict(sorted(dick.items(), key=lambda item: item[1], reverse=True))

    # Создаем новый массив, в котором элементы повторяются в соответствии с их количеством в отсортированном словаре
    sorted_arr = []
    for num, count in sorted_dict.items():
        sorted_arr.extend([num] * count)

    # Выводим отсортированный массив
    # print(sorted_arr)
    count = 0
    vvv = {}
    for i in set(numbers):
        # print(i)
        if dick[i] >= k:
            vvv[i] = k - 1
            del dick[i]
    if len(vvv) == 0:
        print(len(numbers))
    else:
        count = sum(vvv.values())
        for i in dick:
            if dick[i] + count >= k:
                count = k - 1
        print(count)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
trials = int(input())

for testCase in range(trials):

  n, k = map(int, input().split())
  intList = [int(x) for x in input().split()]

  if k > n:
    print(n)
  else:
    dic = {}

    for i in intList:
      if i in dic:
        dic[i] += 1
      else:
        dic[i] = 1

    l = [(k, v) for k, v in sorted(dic.items(), reverse=True, key=lambda item: item[1])]

    newList = [item for tuple in l for item in [tuple[0]] * tuple[1]]

    pointer = 0
    c = 0
    while newList[pointer] == newList[pointer + (k - 1)]:
      for i in range(k - 1):
        newList[pointer + (1 + i)] = newList[pointer + k]

      pointer += 1
      c += 1
      if pointer + k >= n:
        break

    if c == 0:
      print(n)
    else:
      print(n - (c + 1))

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given code, we need to focus on scenarios where the code might fail. The code seems to handle the basic cases but might have issues with more complex scenarios, especially those involving the sorting and the loop logic.

### Step-by-Step Analysis:
1. **Edge Case: k > n**:
   - The code correctly handles this case by printing `n`.
   - Example: `1 2 1 2` (1 test case, n=2, k=1)

2. **Single Unique Card**:
   - If all cards are unique, no operations can be performed.
   - Example: `1 5 1 2 3 4 5`

3. **Multiple Identical Cards**:
   - If there are multiple identical cards, the code should handle the exchange correctly.
   - Example: `1 10 2 2 2 2 2 2 2 2 2 2`

4. **Mixed Cards**:
   - If there are a mix of cards with some having the same number and others being unique, the code might fail if it doesn't handle the sorting and loop logic correctly.
   - Example: `1 7 2 3 2 3 2 3 2`

5. **Edge Case: k = 2**:
   - When k = 2, the code should handle the simplest case of exchanging pairs.
   - Example: `1 6 1 1 2 2 3 3`

6. **Edge Case: k = n**:
   - When k equals n, the code should handle the case of exchanging all cards at once.
   - Example: `1 5 1 1 1 1 1`

### Chosen Test Input:
Let's choose a test input that involves a mix of cards with some having the same number and others being unique, which can expose issues in the sorting and loop logic.

```plaintext
1 7 2 3 2 3 2 3 2
```

This input will test the code's ability to handle a mix of cards and ensure that the loop logic and sorting are working correctly. If the code fails to handle this case, it will likely fail in more complex scenarios as well.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
public class Solution {
	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, Integer> map = new HashMap<>();
			int s=0;
			for(int i=0; i<n; i++) {
				int m =scanner.nextInt();
				if(map.containsKey(m) && map.get(m)<k-1 ) {
					map.put(m, map.get(m)+1);
				}else if(map.containsKey(m) && map.get(m)==k-1){
					map.remove(m);
					s+=k-1;
				}
				else {
					map.put(m, 1);
				}
			}

			if (map.isEmpty()) {
				System.out.println(s);
			}else {
			List<Integer> sortedValues = new ArrayList<>(map.values());
		    Collections.sort(sortedValues, Collections.reverseOrder());
		    int res=0;
			for ( int v : sortedValues) {
				if ( s-k+v>=0) {
					s-=k-v;
					s+=k-1;
				}else {
					res+=v;
				}

			}
			res+=(s/k)*(k-1);
			res = (res < k - 1 && k<n) ? k - 1 : res;
			System.out.println(res);
			}
			}

		}

	}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that might cause the code to fail. Here are some considerations:

1. **Cases with a small number of cards** (e.g., \( n = k \)) where all cards have the same number.
2. **Cases where \( k \) is larger than the number of unique cards**.
3. **Cases with a large number of unique cards**.
4. **Cases where the number of cards is just one more than a multiple of \( k \)**.

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

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

This test input includes:
1. A case where \( n = k \) and all cards have the same number.
2. A case where \( k \) is larger than the number of unique cards.
3. A case with multiple sets of \( k \) cards with the same number.
4. A case where the number of cards is just one more than a multiple of \( k \).
5. A case with a large number of cards that can be exchanged multiple times.

These scenarios should help expose any issues 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: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
public class Solution {
	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, Integer> map = new HashMap<>();
			int s=0;
			for(int i=0; i<n; i++) {
				int m =scanner.nextInt();
				if(map.containsKey(m) && map.get(m)<k-1 ) {
					map.put(m, map.get(m)+1);
				}else if(map.containsKey(m) && map.get(m)==k-1){
					map.remove(m);
					s+=k-1;
				}
				else {
					map.put(m, 1);
				}
			}

			List<Integer> sortedValues = new ArrayList<>(map.values());
		    Collections.sort(sortedValues, Collections.reverseOrder());
		    int res=0;
			for ( int v : sortedValues) {
				if ( s-k+v>=0) {
					s-=k-v;
					s+=k-1;
				}else {
					res+=v;
				}

			}
			res+=(s/k)*(k-1);
			res = (res < k - 1 && k<=n) ? k - 1 : res;
			System.out.println(+res);
			}

		}

	}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the steps to think through the process:

1. **Understand the Bug**: The buggy code has issues with how it handles the removal and re-addition of cards. Specifically, it may not correctly account for the number of cards removed and added back during the exchange operations.

2. **Edge Cases**:
   - **Small `n` and `k`**: Test cases where `n` and `k` are small to ensure the algorithm handles the basic operations correctly.
   - **Large `n` and `k`**: Test cases where `n` and `k` are large to check if the algorithm can handle large inputs efficiently.
   - **Multiple cards with the same number**: Test cases where there are multiple cards with the same number to check if the algorithm correctly performs the exchange operations.
   - **No possible exchanges**: Test cases where no exchange is possible to ensure the algorithm handles such cases correctly.

3. **Challenging Scenarios**:
   - **Exact multiples of `k`**: Test cases where the number of cards with a specific number is an exact multiple of `k`.
   - **Just below multiple of `k`**: Test cases where the number of cards with a specific number is just below a multiple of `k`.
   - **Just above multiple of `k`**: Test cases where the number of cards with a specific number is just above a multiple of `k`.

4. **Test Input**:
   - **Test Case 1**: A small number of cards with a number that is an exact multiple of `k`.
   - **Test Case 2**: A small number of cards with a number that is just below a multiple of `k`.
   - **Test Case 3**: A small number of cards with a number that is just above a multiple of `k`.
   - **Test Case 4**: A large number of cards with a number that is an exact multiple of `k`.
   - **Test Case 5**: A large number of cards with a number that is just below a multiple of `k`.
   - **Test Case 6**: A large number of cards with a number that is just above a multiple of `k`.

Given these considerations, here is a small-scale test input that includes various challenging scenarios:

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

This test input includes:
1. A case with exactly 3 cards of the same number.
2. A case with 4 cards, 3 of which are the same.
3. A case with 5 cards, 3 of which are the same.
4. A case with 10 cards, all of which are the same.
5. A case with 10 cards, 8 of which are the same and 2 are different.
6. A case with 10 cards, 9 of which are the same and 1 is different.

These cases should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
/*keldr24/172/1/8v539*/
#include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
    int n,k;
    cin>>n>>k;
    vector <int> a(n+1);
    map <int,int> mp;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        mp[a[i]]++;
    }
    vector <int> used;
    for(auto it : mp)
    {
        used.push_back(it.second);
    }
    sort(used.rbegin(),used.rend());
    if(used.size()==1)
    {
        cout<<min(n,k-1)<<endl;
        return;
    }
    for(int i=0;i<used.size();i++)
    {
        if(used[i]<k)
        {
            int ans=0;
            for(int j=i;j<used.size();j++)
            {
                ans+=used[j];
            }
            cout<<ans<<endl;
            return;
        }else
        {
            used[i+1]+=used[i]-1;
            used[i]=0;
        }
    }
    if(used.size()>=k)
    {
        cout<<k-1<<endl;
    }else
    {
        cout<<used.size()<<endl;
    }
}
signed main()
{
    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
3
4 2
1 1 2 2
5 3
1 1 1 2 2
6 3
1 1 1 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
/**
* @Author:your.nemesis
**/

#include <iostream>
#include <string>
#include <utility>
#include <cstdlib>
#include <limits>
#include <cmath>
#include <climits>
#include <vector>
#include <bits/stdc++.h>
#include <set>
#include <map>
#include <iomanip>
#include <unordered_set>
#include <unordered_map>
#include <iterator>
#include <algorithm>
#include <sstream>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//using namespace __gnu_pbds;
using namespace std;

typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;

#define mod 1000000007
#define inf 1e9
#define FIO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define w(t) ll tt;cin>>tt;while(tt--)
#define pb push_back

#define all(arr) arr.begin(),arr.end()

// template <typename T>
// using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
//options: less,greater_equal,less_equal,greater

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}

#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

const int N=3e5+5;
const long long infll = 0x3f3f3f3f3f3f3f3fLL;
int p[N];

ll modulo(ll a, ll b, ll n){
    ll x=1, y=a;
    while (b > 0) {
        if (b%2 == 1) {
            x = ((x%n) * (y%n)) % n; // multiplying with base

        }
        y = ((y%n) * (y%n)) % n; // squaring the base
        b /= 2;
    }
    return x % n;
}

ll modmod(ll a,ll b,ll c){
    // Fermats Little Theorem
    // A^(M-1) = 1 (mod M) if M is a prime.
    // So write B^C as x*(M-1) + y

    ll y = modulo(b,c,mod-1);

    return modulo(a,y,mod);
}

ll value[1000001];
void fillNumberOfDivisorsArray(){
    for(ll i=1;i<=1000000;i++){
        for(ll j=1;i*j<=1000000;j++){
            value[i*j]++;
        }
    }
}

vector<ll> prime;
bool is_composite[N];

void sieve(ll n) {

    fill(is_composite,is_composite + n,false);

    for (ll i=2;i<n;i++){

        if(!is_composite[i]){
            prime.push_back(i);
        }

        for (int j=2;i*j < n;j++){
            is_composite[i * j] = true;
        }
    }
}

void addEdge(vector<ll> adj[],ll u,ll v){
    adj[u].pb(v);
    adj[v].pb(u);
}

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

vector<ll> fact;
void factorialFill(){
    fact.pb(1);
    fact.pb(1);
    for(ll i=2;i<=1000000;i++){
        fact.pb((i%mod * fact.back()%mod)%mod);
    }
}
ll getfact(ll num){
    return fact[num]%mod;
}

ll getBits(ll num){
    return log2(num) + 1;
}

struct cmp {
    bool operator() (const pair<int, int> &a, const pair<int, int> &b) const {
        return a.first < b.first;
    }
};
// void solve(){

// }
// int main(){
//     int t;
//     cin>>t;
//     // t = 1;
//     while(t--){
//         int n;
//         cin>>n;
//         vector<int> a(n);
//         for(int i = 0 ; i < n ; i++){
//             cin>>a[i];
//         }
//         int ans;
//         if(abs(a[n-1] - a[n-2]) < 2){
//             ans = abs(a[n-1] - a[n-2]) + 1;
//         }
//         else{
//             ans = abs(a[n-1] - a[n-2]);
//         }
//         cout<<ans<<endl;

//     }
// }

// #include <iostream>
// #include <vector>
// #include <algorithm>
// using namespace std;

// int main() {
//     int t;
//     cin >> t;
//     while (t--) {
//         int n;
//         cin >> n;
//         vector<int> p(n + 1); // Index 0 is unused to make indexing easier
//         vector<bool> invited(n + 1, false); // Track invited friends
//         int invited_count = 0;

//         // Input friends' best friends and track invited friends
//         for (int i = 1; i <= n; i++) {
//             cin >> p[i];
//             if (invited[p[i]]) {
//                 invited[i] = true;
//                 invited_count++;
//             }
//         }

//         // Check if the count of invited friends is less than 2
//         if (invited_count < 2) {
//             // Invite additional friends to ensure at least two friends attend the party
//             for (int i = 1; i <= n; i++) {
//                 if (!invited[i]) {
//                     invited[i] = true;
//                     invited_count++;
//                     if (invited_count == 2) break; // Stop when at least two friends are invited
//                 }
//             }
//         }

//         cout << invited_count << endl;
//     }
//     return 0;
// }
//Problem 1956/C
// int main(){
//     int t;
//     cin>>t;
//     while(t--){
//         int n;
//         cin>>n;
//         vector<int> a(n);
//         vector<int> b(n);
//         for(int i = 0 ; i < n ; i++){
//             cin>>a[i];
//         }
//         for(int i = 0 ; i < n ; i++){
//             cin>>b[i];
//         }
//         int c = 0;
//         for(int i = 0 ; i < n ; i++){
//             while(a[i] > b[i]){
//                 int w = b[i];
//                 a.pop_back();
//                 a.insert(a.begin() + i,w);
//                 c++;
//             }
//         }
//         cout<<c<<endl;
//     }
// }/

// void solve1(){
//     int n;
//     cin>> n;
//     vector<int> a(n);
//     vector<int> dp(n+1,0);
//     for(int i = 0 ; i < n ; i++){
//         cin>>a[i];
//     }
//     dp[0] = 1;
//     for(int i = 1 ; i < n; i++){
//         dp[i] = dp[i-1]*bool((i+1)-a[i] <= 0);
//     }
//     for(int i = 0 ; i < n ; i++){
//         cout<<dp[i];
//     }
// }

// int gcd(int a, int b)
// {
//     // Find Minimum of a and b
//     int result = min(a, b);
//     while (result > 0) {
//         if (a % result == 0 && b % result == 0) {
//             break;
//         }
//         result--;
//     }

//     // Return gcd of a and b
//     return result;
// }

// void solve1(){
//     int x;
//     cin >> x;
//     vector<int> a(x);
//     a[0] = 0;
//     vector<int> v(x);
//     int ans = 1;
//     for(int i = 1; i < x; i++){
//         a[i] = gcd(x, i) + i;
//     }
//     for(int i = 0 ; i < x ;i++){
//         v[i] = a[i];
//     }
//     sort(v.begin(),v.end(),greater<int>());
//     for(int i = 0 ; i < x ; i++){
//         if(a[i] == v[0]){
//             ans = i;
//         }
//     }
//     cout<<ans<<endl;

// }

// #define MOD 1000000007
// using namespace std;

// void solve1(){
//     int n;
//     cin>> n;
//     vector<int> a(n);
//     vector<int> dp(n + 1);
//     for(int i = 0 ; i < n ; i++){
//         cin>>a[i];
//     }
//     dp[0] = 1;
//     int maxl = 1;
//     for(int i = 1 ; i < n ;i++){
//         dp[i] = (dp[i-1] * (bool(a[i] > a[i-1]))) + 1;
//         maxl = max(maxl,dp[i]);
//     }
//     cout <<maxl<< endl;
// }

// void solve1(){
//     int n;
//     int x;
//     cin>> x >> n;
//     vector<int> a(x);
//     vector<int> dp(n + 1);
//     for(int i = 0 ; i < x ; i++){
//         cin>>a[i];
//     }
//     dp[0] = 1;
//     for(int i = 1 ; i <= n ;i++){
//         for(int j = 0 ; j < x ; j++){
//             if(a[j] <= i){
//                 dp[i] = (dp[i] + dp[i-a[j]]) % MOD;
//             }

//         }
//     }

//     //final subproblem
//     cout <<dp[n]<< endl;
// }

// void solve2(){
//     int n;
//     cin >> n;
//     vector<int> dp(n + 1);
//     // dp[i] = number of ways to make a sum of n
//     // such that the current sum is i

//     // base case
//     dp[n] = 1;
//     for(int i = n - 1; i >= 0; i--){
//         for(int j = 1; j <= 6; j++){
//             if(j <= n - i){
//                 // transition
//                 dp[i] = (dp[i] + dp[i + j]) % MOD;
//             }
//         }
//     }

//     //final subproblem
//     cout << dp[0] << endl;
// }
// int gcd(int a, int b) {
//     if (b == 0)
//         return a;
//     return gcd(b, a % b);
// }
// void solve1(){
//     int x;
//     cin >> x;
//     vector<int> a(x);
//     a[0] = 0;
//     vector<int> v(x);
//     int ans = 1;
//     for(int i = 1; i < x; i++){
//         a[i] = gcd(x, i) + i;
//     }
//     for(int i = 0 ; i < x ;i++){
//         v[i] = a[i];
//     }
//     sort(v.begin(),v.end(),greater<int>());
//     for(int i = 0 ; i < x ; i++){
//         if(a[i] == v[0]){
//             ans = i;
//         }
//     }
//     cout<<ans<<endl;

// }
// int main(){
//     int t;
//     cin>>t;
//     while(t--){
//         solve1();
//     }
// }

// int main() {
//     int t;
//     cin>>t;
//     while(t--){
//         int n;
//         cin>>n;
//         vector<int> a(n);
//         vector<int> x(n-1);
//         for(int i = 0; i < n ; i++){
//             cin>>x[i];
//         }
//         a[0] = x[0] + 1;
//         for(int i =  ; i < n+1 ; i++){
//             x[i] = (a[i-1]%a[i])
//         }
//     }

//     // solve2();
// }

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

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

//         string a, b;
//         cin >> a >> b;

//         int k = 0;
//         int j = 0;
//         for (int i = 0; i < m && j < n; ++i) {
//             if (b[i] == a[j]) {
//                 ++k;
//                 ++j;
//             }
//         }

//         cout << k << endl;
//     }

//     return 0;
// }

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

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

//         vector<int> x(n - 1);
//         for (int i = 0; i < n - 1; ++i) {
//             cin >> x[i];
//         }

//         ll a1 = x[0] + 1;
//         cout << a1 << " ";
//         ll a = a1;
//         for(int i = 0; i < n-1 ; i++){
//             a += x[i];
//             cout << a << " ";
//         }

//         cout << endl;
//     }

//     return 0;
// }

// using namespace std;
// int main(){
// 	int t;
// 	cin>>t;
// 	while(t--){
// 		int n;
// 		cin>>n;
// 		int S=1000;
// 		cout<<S<<" ";
// 		for(int i=2;i<=n;i++){
// 			int x;
// 			cin>>x;
// 			S+=x;
// 			cout<<S<<" ";
// 		}
// 		cout<<"\n";
// 	}
// }

// int main(){
//     string s;
//     cin>>s;
//     int q = 0 , a = 0;
//     for(int i = 0 ; i < s.size() ; i++){
//         if(s[i] == 'Q'){
//             q++;
//         }
//         if(s[i] == 'A'){
//             a++;
//         }
//     }
//     if(q%2 && a == 1){
//         q = q/2;
//     }
// }

// int main() {
//     int t;
//     cin >> t;
//     while (t--) {
//         int n;
//         cin >> n;
//         vector<int> a(n);
//         for (int i = 0; i < n; i++) {
//             cin >> a[i];
//         }
//         sort(a.begin(), a.end());
//         int c = 0;
//         vector<int> freq(a.back() + 1);
//         for (int i = 0; i < n; i++) {
//             freq[a[i]]++;
//         }
//         for (int i = 0; i <= a.back(); i++) {
//             if (freq[i] >= 3) {
//                 c += freq[i] / 3;
//             }
//         }
//         cout << c << endl;
//     }
//     return 0;
// }

int main(){
    int t;
    cin>>t;
    while(t--){
        int n,k;
        cin>> n >> k;
        bool f = false;
        vector<int> a(n);
        vector<int> freq(101,0);
        for(int i = 0 ; i < n ; i++){
            cin>>a[i];
        }
        if(n == 1){
            cout<<1<<endl;
            continue;
        }
        for(int i = 0 ; i < n ; i++){
            freq[a[i]]++;
        }
        sort(freq.begin(),freq.end());
        for(int i = 0 ; i < freq.size() ; i++){
            if(freq[i] >= k || freq[i] == n){
                f = true;
                break;
            }
        }
        if(f){
            cout<<k-1<<endl;
        }
        else{
            cout<<n<<endl;

        }

    }
}

// const int INF = 1'000'000'009;
// int t;

// char type(const vector <int>& a, int id) {
//     int distL = (id == 0? INF : a[id] - a[id - 1]);
//     int distR = (id + 1 == a.size()? INF : a[id + 1] - a[id]);
//     if(distL < distR) return 'L';
//     if(distL > distR) return 'R';
//     assert(false);
// }

// int main() {

//     cin >> t;
//     for (int tc = 0; tc < t; ++tc) {
//         int n;
//         cin >> n;
//         vector <int> a(n);
//         for (int i = 0; i < n; ++i)
//             cin >> a[i];

//         vector <int> l(n), r(n);
//         for (int i = 1; i < n; ++i)
//             r[i] = r[i - 1] + (type(a, i - 1) == 'R'? 1 : a[i] - a[i - 1]);
//         for (int i = n - 2; i >= 0; --i)
//             l[i] = l[i + 1] + (type(a, i + 1) == 'L'? 1 : a[i + 1] - a[i]);

//         int m;
//         cin >> m;
//         for (int i = 0; i < m; ++i) {
//             int x, y;
//             cin >> x >> y;
//             --x, --y;
//             if (x < y)
//                 cout << r[y] - r[x] << endl;
//             else
//                 cout << l[y] - l[x] << 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 consider the following:

1. **Edge Cases:**
   - Single card scenario (`n = 1`).
   - Scenario where all cards have the same number (`n = k`).
   - Scenario where the number of cards is just less than `k` (`n = k - 1`).
   - Scenario where the number of cards is just more than `k` (`n = k + 1`).
   - Scenario with a mix of different numbers.

2. **Critical Scenarios:**
   - When `k` is equal to `n` and all cards have the same number.
   - When `k` is equal to `n` and cards have different numbers.
   - When `k` is less than `n` and there are multiple cards with the same number.

3. **Buggy Code Analysis:**
   - The buggy code checks if any frequency is greater than or equal to `k` and prints `k-1` if true.
   - It prints `n` if no frequency meets the condition.
   - This logic is flawed because it does not consider the possibility of multiple operations to reduce the number of cards.

Given this analysis, let's create a test input that will expose the bug:

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

Explanation:
1. **Test Case 1:**
   - `n = 3`, `k = 3`
   - All cards have the same number (`1 1 1`).
   - Expected output: `2` (since we can exchange 3 cards for 2 cards).

2. **Test Case 2:**
   - `n = 4`, `k = 2`
   - All cards have different numbers (`1 2 3 4`).
   - Expected output: `2` (since we can exchange 2 cards for 1 card, and we can do this twice).

3. **Test Case 3:**
   - `n = 5`, `k = 3`
   - Some cards have the same number (`1 1 1 2 2`).
   - Expected output: `2` (since we can exchange 3 cards with the same number for 2 cards, and we are left with 2 cards).

4. **Test Case 4:**
   - `n = 6`, `k = 3`
   - All cards have the same number (`1 1 1 1 1 1`).
   - Expected output: `2` (since we can exchange 3 cards for 2 cards, and we can do this twice).

These test cases cover a range of scenarios that should expose the logic flaws in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

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

    arr = [*map(int, input().split())]
    c = dict()
    for i in arr:
        c[i] = c.get(i, 0) + 1
    vals = sorted(c.values())
    vals = vals[::-1]
    extra = 0
    changes = 1
    while changes:
        changes = 0
        vals.sort(reverse=True)
        for i in range(len(vals)):
            v = vals[i]
            if v < k: continue
            u, w = divmod(v, k)
            vals[i] = w
            if u:
                if (i+1) == len(vals):
                    vals.append(u * (k-1))
                else:
                    Q, I = 0, 0
                    for t, r in enumerate(vals):
                        if r > Q:
                            Q = r
                            I = t
                    vals[I] += u * (k-1)
                changes = True
                break
    print(sum(vals))

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

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

Think step by step.

Response:
```plaintext
1
10 3
1 1 1 1 1 1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
t = int(input())
for _ in range(t):
    n,k=map(int,input().split())
    arr=list(map(int,input().split()))
    max_count = 0
    element = 0
    for x in arr:
        count = arr.count(x)
        if count > max_count:
            max_count = count
            element = x
    if(k>max_count):
        print(n)
    elif(k==max_count):
        if(max_count%2!=0):
            print(2)
        else:
            print(1)
    else:
        print(k-1)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class A {
    private static InputReader in;
    private static PrintWriter out;

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

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

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

    static final int mod = 1_000_000_007;

    public static void main(String[] args) {
        in = new InputReader(System.in);
        out = new PrintWriter(System.out);
        Task solver = new Task();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve();
        out.close();
    }

    static class Task {
        public void solve() {
            int n = in.nextInt(), k = in.nextInt();
            int[] a = in.readArray(n);
            int[] c = new int[101];
            for (int i = 0; i < n; i++) {
                c[a[i]]++;
            }
            int ans = 0;
            for (int i = 0; i < 101; i++) {
                if (c[i] >= k) {
                    ans += k - 1;
                }
            }
            int v = 0;
            for (int i = 0; i < 101; i++) {
                if (c[i] >= k) continue;
                if (c[i] < k && ans >= k - c[i]) {
                    ans -= k - c[i];
                    ans += k - 1;
                }
                else v += c[i];
            }
            out.println(ans + v);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Exchange
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a hand of $$$n$$$ cards, where each card has a number written on it, and a fixed integer $$$k$$$. You can perform the following operation any number of times:

- Choose any $$$k$$$ cards from your hand that all have the same number.
- Exchange these cards for $$$k-1$$$ cards, each of which can have any number you choose (including the number written on the cards you just exchanged).

Here is one possible sequence of operations for the first example case, which has $$$k=3$$$:

What is the minimum number of cards you can have in your hand at the end of this process?

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$2 \le k \le 100$$$) — the number of cards you have, and the number of cards you exchange during each operation, respectively.

The next line of each test case contains $$$n$$$ integers $$$c_1, c_2, \ldots c_n$$$ ($$$1 \le c_i \le 100$$$) — the numbers written on your cards.

Output: For each test case, output a single integer — the minimum number of cards you can have left in your hand after any number of operations.

Examples:
input:
7
5 3
4 1 1 4 4
1 10
7
7 2
4 2 1 100 5 2 3
10 4
1 1 1 1 1 1 1 1 1 1
5 2
3 8 1 48 7
6 2
10 20 30 10 20 40
6 3
10 20 30 10 20 40
output:
2
1
1
3
5
1
6

Note: The first example case corresponds to the picture above. The sequence of operations displayed there is optimal, so the answer is $$$2$$$.

In the second example case, no operations can be performed, so the answer is $$$1$$$.

In the fourth example case, you can repeatedly select $$$4$$$ cards numbered with $$$1$$$ and replace them with $$$3$$$ cards numbered with $$$1$$$, until there are $$$3$$$ cards left.

Buggy Code:
import java.util.*;

public class Main {
    static final int MOD = 1000000007;

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

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

            int[] a = new int[n];

            int[] temp = new int[1000001];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                temp[a[i]]++;
            }

            boolean flag = false;
            for (int i = 0; i < 1000001; i++) {
                if (temp[i] >= k) {
                    flag = true;
                    break;
                }
            }

            if (!flag) {
                System.out.println(n);
            } else {
                System.out.println(k - 1);
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
7 3
1 1 1 2 2 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize the Difference
time_limit_ms: 2000
memory_limit_mb: 256
Description: Zhan, tired after the contest, gave the only task that he did not solve during the contest to his friend, Sungat. However, he could not solve it either, so we ask you to try to solve this problem.

You are given an array $a_1, a_2, \ldots, a_n$ of length $n$. We can perform any number (possibly, zero) of operations on the array.

In one operation, we choose a position $i$ ($1 \leq i \leq n - 1$) and perform the following action:

  * $a_i := a_i - 1$, and $a_{i+1} := a_{i+1} + 1$.

Find the minimum possible value of $\max(a_1, a_2, \ldots, a_n) - \min(a_1, a_2, \ldots, a_n)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^{12}$).

The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the minimum possible value of $\max(a_1, a_2, \ldots, a_n) - \min(a_1, a_2, \ldots, a_n)$.

Examples:
input:
5
1
1
3
1 2 3
4
4 1 2 3
4
4 2 3 1
5
5 14 4 10 2
output:
0
2
1
1
3

Note: In the third testcase, you can perform the operation twice with $i = 1$.

After that, the array is $a = [2, 3, 2, 3]$, and $\max(2, 3, 2, 3) - \min(2, 3, 2, 3) = 3 - 2 = 1$.

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

using namespace std;

typedef long long int ll;

ll ceil2(ll a, ll b) {
    if (a == 0) return 0;
    return (a - 1)/b + 1;
}

void solve()
{
    int n;
    cin>>n;

    vector<ll> A(n);

    ll sumright=0;

    for(int i=0;i<n;i++)
    {
        cin>>A[i];
        if(i>0)
        {
            sumright += A[i];
        }
    }
    //cout << sumright << endl;
    if(n==1)
    {
        cout << 0 << endl;
        return;
    }
    ll divaftersum = sumright/(n-1);
    ll ceilval=ceil2(sumright,n-1);
    ll floorval=sumright/(n-1);
    //cout << A[0] << " " << floorval << " " << ceilval << " " << endl;

    while(A[0]>divaftersum)
    {
        --A[0];
        ++sumright;
        divaftersum = sumright/(n-1);
        ceilval=ceil2(sumright,n-1);
        floorval=sumright/(n-1);
        //cout << A[0] << " " << floorval << " " << ceilval << " " << endl;
    }
    cout << max(ceilval,A[0])- min(A[0],floorval)<< endl;

}

int main()
{
    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
3
1
1
2
1 1
3
1 1000000000000 1
```