**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cat, Fox and Maximum Array Split
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is an interactive problem.

Fox gave Cat two positive integers $$$n$$$ and $$$k$$$. She has a hidden array $$$a_1, \ldots , a_n$$$ of length $$$n$$$, such that $$$1 \leq a_i \leq n$$$ for every $$$i$$$. Now they are going to play the following game:

For any two integers $$$l, r$$$ such that $$$1 \leq l \leq r \leq n$$$, define $$$f(l, r) = (r - l + 1) \cdot \max\limits_{x=l}^r a_x$$$. In other words, $$$f(l, r)$$$ is equal to the maximum of the subarray $$$a_l, \ldots, a_r$$$ multiplied by its size.

Cat can ask Fox at most $$$2 n$$$ questions about the array. He will tell her two integers $$$l$$$ and $$$x$$$ ($$$1 \leq l \leq n, 1 \leq x \leq 10^9$$$), and she will tell him one integer $$$p$$$ as the answer — the smallest positive integer $$$r$$$ such that $$$f(l, r) = x$$$, or $$$n+1$$$ if no such $$$r$$$ exists.

Now, Cat needs to find the largest value $$$m$$$ such that there exists a sequence $$$c_1, \ldots, c_{k-1}$$$ such that $$$1 \leq c_1 < \ldots < c_{k-1} < n$$$ and $$$f(1, c_1) = f(c_1 + 1, c_2) = \ldots = f(c_{k-1}+1, n) = m$$$. If no such $$$m$$$ exists, he should indicate this and take $$$-1$$$ as the answer. Note that for $$$k = 1$$$, $$$m$$$ is always equal to $$$f(1, n)$$$.

In other words, the goal is to find the largest $$$m$$$ such that you can split the array into exactly $$$k$$$ subarrays ($$$k$$$ is the constant given to you in the beginning of the interaction) so that all the subarrays have the product of their length and their maximum equal to $$$m$$$, or determine that no such $$$m$$$ exists. Every element should belong in exactly one of the subarrays.

Cat doesn't know what he should do, so he asked you to play the game for him.

Input: None

Output: None

Examples:
input:
3
1 1

1
2 2

1

3

1
6 3

7

2

3

6

1
output:
! 1

? 1 1

? 2 1

! -1

? 1 9

? 1 6

? 3 6

? 4 6

! 6

Note: The hidden arrays in the three testcases are $$$[1]$$$, $$$[1, 2]$$$ and $$$[1, 3, 6, 1, 2, 1]$$$. In the second testcase, no split satisfies the constraints, so the answer is $$$-1$$$.

The answer for the first query of the third testcase is $$$7$$$ since no valid $$$r$$$ exists. For the second query of the third testcase, since $$$2 \cdot \max(1, 3) = 6$$$, we will get $$$2$$$ as the answer, since $$$r = 1$$$ doesn't satisfy the constraint.

The sample interaction guessed all three answers ($$$1, -1$$$ and $$$6$$$) correctly, so it received $$$1$$$ after each answer.

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

public class D {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);

	static int count = 0;

	public static void main(String[] args) throws IOException {

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = sc.nextInt();

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void solve(int t) {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			int n = sc.nextInt();
			int k = sc.nextInt();

			int max = n;

			count = 2*n;

			while (query(1, max * n) == -1) {
				--max;
			}

			int maxUse = n / k;
			int target;

			int val;

			int result = -1;

			for (int i = maxUse; i >= 1; --i) {
				val = solve(n, i, k, max);

				if (val == -1) {
					break;
				}else if (val == 0) {
					result = max * i;
					break;
				}
			}

			System.out.println("! " + result);
			System.out.flush();
			sc.nextInt();
		}

		private int solve(int n, int length, int k, int max) {

			int currentPos = 1;
			int target = length * max;

			int end = 0;

			int minEnd;

			long maxEnd;

			for (int i = 1, left = k - 1; i <= k; ++i, --left) {
				end = query(currentPos, target);

				if (end == n + 1) {
					return 1;
				}

				currentPos = end + 1;
				minEnd = left * length + end;

				// System.out.println(minEnd);

				if (minEnd > n) {
					return 1;
				}

				maxEnd = left;
				maxEnd *= max;
				maxEnd += end;

				if (maxEnd < n) {
					return -1;
				}
			}
			// [1,3,6,1,2,1]

			return 0;
		}

		private int query(int l, int x) {
			if (count-- < 0) {
				throw new RuntimeException();
			}

			System.out.println("? " + l + " " + x);
			System.out.flush();

			return sc.nextInt();
		}
	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

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

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

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

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

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

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

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

   }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cat, Fox and Maximum Array Split
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is an interactive problem.

Fox gave Cat two positive integers $$$n$$$ and $$$k$$$. She has a hidden array $$$a_1, \ldots , a_n$$$ of length $$$n$$$, such that $$$1 \leq a_i \leq n$$$ for every $$$i$$$. Now they are going to play the following game:

For any two integers $$$l, r$$$ such that $$$1 \leq l \leq r \leq n$$$, define $$$f(l, r) = (r - l + 1) \cdot \max\limits_{x=l}^r a_x$$$. In other words, $$$f(l, r)$$$ is equal to the maximum of the subarray $$$a_l, \ldots, a_r$$$ multiplied by its size.

Cat can ask Fox at most $$$2 n$$$ questions about the array. He will tell her two integers $$$l$$$ and $$$x$$$ ($$$1 \leq l \leq n, 1 \leq x \leq 10^9$$$), and she will tell him one integer $$$p$$$ as the answer — the smallest positive integer $$$r$$$ such that $$$f(l, r) = x$$$, or $$$n+1$$$ if no such $$$r$$$ exists.

Now, Cat needs to find the largest value $$$m$$$ such that there exists a sequence $$$c_1, \ldots, c_{k-1}$$$ such that $$$1 \leq c_1 < \ldots < c_{k-1} < n$$$ and $$$f(1, c_1) = f(c_1 + 1, c_2) = \ldots = f(c_{k-1}+1, n) = m$$$. If no such $$$m$$$ exists, he should indicate this and take $$$-1$$$ as the answer. Note that for $$$k = 1$$$, $$$m$$$ is always equal to $$$f(1, n)$$$.

In other words, the goal is to find the largest $$$m$$$ such that you can split the array into exactly $$$k$$$ subarrays ($$$k$$$ is the constant given to you in the beginning of the interaction) so that all the subarrays have the product of their length and their maximum equal to $$$m$$$, or determine that no such $$$m$$$ exists. Every element should belong in exactly one of the subarrays.

Cat doesn't know what he should do, so he asked you to play the game for him.

Input: None

Output: None

Examples:
input:
3
1 1

1
2 2

1

3

1
6 3

7

2

3

6

1
output:
! 1

? 1 1

? 2 1

! -1

? 1 9

? 1 6

? 3 6

? 4 6

! 6

Note: The hidden arrays in the three testcases are $$$[1]$$$, $$$[1, 2]$$$ and $$$[1, 3, 6, 1, 2, 1]$$$. In the second testcase, no split satisfies the constraints, so the answer is $$$-1$$$.

The answer for the first query of the third testcase is $$$7$$$ since no valid $$$r$$$ exists. For the second query of the third testcase, since $$$2 \cdot \max(1, 3) = 6$$$, we will get $$$2$$$ as the answer, since $$$r = 1$$$ doesn't satisfy the constraint.

The sample interaction guessed all three answers ($$$1, -1$$$ and $$$6$$$) correctly, so it received $$$1$$$ after each answer.

Buggy Code:
// AUTHOR: UNIVERSAL ADMIN

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

public class Main {
    public static void main(String[] args) {
        new Thread(null, ()-> {
            new Main2().start();
        }, "", 1 << 27).start();
    }
}
//----------------------------------------------------------------------------------------------------------------------------------------------
class Main2 {
    void start() {
        int t = ni();
        while(t-- > 0)
            solve();

        w.close();
    }

    int askCt = 0;
    int ask (int l, int x) {
        askCt++;
        p("? " + l + " " + x);
        return ni();
    }

    void solve() {
        askCt = 0;
        int n = ni(), k = ni();
        int nn = n;
        while(true) {
            int l = 1;
            if(askCt == 2*n) {
                break;
            }
            int ans = ask(l, nn);
            l = ans+1;
            int ct = 0;
            while(ans < n) {
                if(askCt == 2*n) {
                    break;
                }
                ans = ask(l, nn);
                l = ans+1;
                ct++;
            }
            ct++;
            if(ct == k && l == n+1) {
                p("! " + nn);
                return;
            }
            nn += n;
        }
        p("! " + -1);
    }
//----------------------------------------------------------------------------------------------------------------------------------------------

    //--------------------------------------------------------------------------------------
    long lma = Long.MAX_VALUE, lmi = Long.MIN_VALUE;
    int ima = Integer.MAX_VALUE, imi = Integer.MIN_VALUE;
    long mod = 1000000007;
    //    long mod = 998244353;

    //--------------------------------------------------------------------------------------
    PrintWriter w = new PrintWriter(System.out, true);
    void p(int i) {w.println(i);} void p(long l) {w.println(l);}
    void p(double d) {w.println(d);} void p(String s) { w.println(s);}
    void pr(int i) {w.print(i);} void pr(long l) {w.print(l);}
    void pr(double d) {w.print(d);} void pr(String s) { w.print(s);}
    void pl() {w.println();}

    //--------------------------------------------------------------------------------------
    public BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");
    String next() {
        while (!st.hasMoreTokens()) {
            try { st = new StringTokenizer(br.readLine());} catch (IOException e) { e.printStackTrace(); }
        }
        return st.nextToken();
    }

    int ni() { return Integer.parseInt(next()); }
    long nl() { return Long.parseLong(next()); }
    double nd() { return Double.parseDouble(next()); }
    String ns() { return next(); }

    int[] na(long n) {int[]ret=new int[(int)n]; for(int i=0;i<n;i++) ret[i]=ni(); return ret;}
    long[] nal(long n) {long[]ret=new long[(int)n]; for(int i=0;i<n;i++) ret[i]=nl(); return ret;}
    Integer[] nA(long n) {Integer[]ret=new Integer[(int)n]; for(int i=0;i<n;i++) ret[i]=ni(); return ret;}
    Long[] nAl(long n) {Long[]ret=new Long[(int)n]; for(int i=0;i<n;i++) ret[i]=nl(); return ret;}
}

**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
6 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
#include <iostream>

using namespace std;
int s,n,k,no_op;
int main()
{
    cin>> n;
    while (n--)
    {
        cin >> s >> k;
        if(s == 1)
        {
            cout << 0 << endl;
        }
        else
        {
            no_op = s/k + s%k;
            cout << no_op+1 <<endl;
        }
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

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

using namespace std;
const int N=1e6+10;
typedef long long ll;
int n,x,m;
int main()
{
    int t;
    cin>>t;
    while(t--){
        cin>>n>>m;
        if(n==1) {
            cout<<0<<endl;
            continue;
        }
        int op=0;
        if(n%m) op=1;
        cout<<n/m+op+1<<endl;
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

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

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

    if(n == 1){
        cout<<0<<"\n";
        return;
    }
    else if(n>1 && n<=k){
        cout<<1<<"\n";
        return;
    }

    int num = n;

    int ans = 0;
    while(num>k){
        ans++;
        int mov = k-1;
        while(num>k && mov--){
            num-=k;
            ans++;
        }
    }
    if(num>1 && num<=k){
        ans++;
    }
    cout<<ans<<"\n";
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int t;
    cin>>t;
    while(t--) solve();

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
def main():
    t = int(input())
    for _ in range(t):
        #n = int(input())
        n,k = map(int,input().split())
        if n==1:
            print(0)
            continue
        if k<n:
            print(n//k + 1 + n%k)
        else:
            print(1)
        #a = list(map(int,input().split()))
main()

**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 2
2 2
1000 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
# python cf.py < cf.txt
# https://stackoverflow.com/questions/59060237/how-to-setup-visual-studio-code-stdin-stdout-redirection-for-python-debugger
# https://stackoverflow.com/a/48650929
# запускай отладку alt+q enter

# одно число
# num: int = int(input())

# числа в одной строке
# string: list[str] = open(0).read().split()
# n, k = map(int, string)
# nums: list[int] = list(map(int, strings[str_pos].split()))

# строки
strings: list[str] = open(0).read().split('\n')

# одна строка
# string: str = open(0).read().split('\n')[0]

##################### imports

# from collections import Counter
import math
################################ РЕШЕНИЕ ####################

test_count: int = int(strings[0])

str_pos: int = 1

result: list[str] = []

for _ in range(test_count):

    # nums: list[int] = list(map(int, strings[str_pos].split()))
    # nums: str = strings[str_pos]
    n, k = map(int, strings[str_pos].split())

    # result.append(str(math.ceil(n / k)))
    # r: int = math.floor((n - k) / (k - 1)) + 1

    part_count: int = math.ceil(n / k)

    if part_count > 1:
        steps_to_decompose: int = math.ceil(part_count / k)
    else:
        steps_to_decompose: int = 0

    r: int = steps_to_decompose + part_count
    # r: int = part_count
    if n % k == 1:
        r -= 1
    elif n < k * part_count and k * part_count - n  <= k - part_count and r > 1:
        r -= 1
    result.append(str(r))
    result.append('\n')

    str_pos += 1

    # if one_count > zero_count:
    #     result.append("YES\n")
    # else:
    #     result.append("NO\n")

    # str_pos += 1

################################ ВЫВОД ####################

# print(' '.join(map(str, A)))

# список строк
# print(*result)
print(''.join(result))

# одно число
# print(result)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
def min_operations(n, k):
    operations = 0
    while n > 1:
        if n % k == 0:
            operations += (n // k)+1
            n = 1
        else:
            operations += (n // k + 1)+1
            n = n % k
    return operations

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    print(min_operations(n, 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
4
1 5
2 2
7 2
15 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
import java.util.Scanner;

public class SplitMultiset {
  private static int solve(int n, int k){
    return (n-1)/(k-1) + (n-1)%(k-1);
  }
  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 res = solve(n,k);
      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:
```plaintext
3
1 2
2 2
3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
import java.util.*;
public class program{
	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 count=0;
			boolean b=true;
			 if(n==1) {
				System.out.println(0);
				b=false;
			}
			 else if(k>=n) {
				System.out.println(1);
				b=false;
			}

			else {

				while(n>k) {
					count++;
					if((n/k)<k||n%k==0) {
						count=count+(n/k);
						break;
					}
					else {
						n=n-(k*(k-1));
						count=count+k-1;
					}

				}
			}
			if(b) {

					System.out.println(count);

			}

		}
	}
}

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

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

Think step by step.

Response:
```plaintext
4
1 5
5 2
6 3
15 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
#include<bits/stdc++.h>
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma comment(linker, "/stack:200000000")
using namespace std;
#define F first
#define S second
#define EB emplace_back
#define MP make_pair
#define all(o) (o).begin(), (o).end()
#define mset(m,v) memset(m,v,sizeof(m))
#define fr(i,n) for(lli i=0;i<(n);++i)
#define rep(i,a,b) for(lli i=a;i<=b;++i)
#define per(i,a,b) for(lli i=a;i>=b;i--)
#define remin(a,b) (a=min((a),(b)))
#define remax(a,b) (a=max((a),(b)))
#define sz(x) (lli)(x).size()
#define endl '\n'
typedef int lli;        typedef long double ld;
typedef pair<lli,lli> ii;     typedef vector<lli> vi;
typedef vector<ii> vii;       typedef vector<vi> graph;
long long MOD=1000000007;     long double EPS=1e-9;  typedef long long ll;
#ifndef ONLINE_JUDGE
#define debarr(a,n)cerr<<#a<<":";for(int i=0;i<n;i++)cerr<<a[i]<<" ";cerr<<endl;
#define debmat(mat,row,col)cerr<<#mat<<":\n";for(int i=0;i<row;i++){for(int j=0;j<col;j++)cerr<<mat[i][j]<<" ";cerr<<endl;}
#define pr(...)dbs(#__VA_ARGS__,__VA_ARGS__)
template<class S,class T>ostream &operator<<(ostream &os,const pair<S,T> &p){return os<<"("<<p.first<<","<<p.second<<")";}
template<class T>ostream &operator<<(ostream &os,const vector<T> &p){os<<"[";for(auto&it:p)os<<it<<" ";return os<<"]";}
template<class T>ostream &operator<<(ostream &os,const set<T>&p){os<<"[";for(auto&it:p)os<<it<<" ";return os<<"]";}
template<class T>ostream &operator<<(ostream &os,const multiset<T>&p){os<<"[";for(auto&it:p)os<<it<<" ";return os<<"]";}
template<class S,class T>ostream &operator<<(ostream &os,const map<S,T>&p){os<<"[";for(auto&it:p)os<<it<<" ";return os<<"]";}
template<class T>void dbs(string str,T t){cerr<<str<<":"<<t<<"\n";}
template<class T,class...S>void dbs(string str,T t,S... s){int idx=str.find(',');cerr<<str.substr(0,idx)<<":"<<t<<",";dbs(str.substr(idx+1),s...);}
#else
#define pr(...){}
#define debarr(a,n){}
#define debmat(mat,row,col){}
#endif
lli binpow(lli b,lli p,lli mod){lli ans=1;b%=mod;for(;p;p>>=1){if(p&1)ans=ans*b%mod;b=b*b%mod;}return ans;}

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

    if(n==1){
        cout<<0<<endl;
        return;
    }

    if(n%k==0){
        cout<<(n/k)+1<<endl;
        return;
    }

    int ans = 0;

    while(n>0){
        n-=k;
        if(n==1){
            ans+=2;
            break;
        }
        ans+=2;
    }

    cout<<ans<<endl;
}

signed main(){
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    //cout<<fixed<<setprecision(15);
    //clock_t begin = clock();
    srand(time(0));lli _t=1;
    cin>>_t;
    for(lli i=1;i<=_t;i++){
        //cout<<"Case #"<<i<<": ";
        solve();
    }
    // clock_t end = clock();double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;cerr << elapsed_secs;
}

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

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

Think step by step.

Response:
```plaintext
4
1 5
5 2
6 3
7 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <vector>

#define int long long
using namespace std;
typedef pair<int, int> pii;

const int N = 200010;

int w[N];

void solve()
{
    int n, k; cin >> n >> k;
    if(k == 2){
        cout <<  n - 1 << endl;
        return ;
    }
    cout << ((n - 1) + 1) / (k - 1) << endl;
}

signed main()
{
    int _; cin >> _;
    while(_--) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
1 5
5 2
6 3
15 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
def mi(a,b):
    if(a==1):
        return 0
    op=0
    while(a>1):
        op+=1
        if(a<=b):
            op+=a-1
            break
        a=(a+b-1)//b
    return op
x=int(input())
r=[]
for i in range(x):
    a,b=map(int,input().split())
    r.append(mi(a,b))
for u in r:
    print(u)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the code and identify potential edge cases or scenarios where the code might fail.

### Code Analysis
1. **Base Case:**
   - If `a == 1`, the function returns `0` immediately. This is correct because if the multiset already contains only one, no operations are needed.
2. **Main Loop:**
   - The loop continues as long as `a > 1`.
   - The number of operations `op` is incremented by 1 for each iteration.
   - If `a <= b`, the code adds `a - 1` to `op` and breaks the loop. This is because if `a` is less than or equal to `b`, you can split `a` into `a` ones in `a - 1` operations.
   - Otherwise, `a` is updated to `(a + b - 1) // b`, which is the ceiling of `a / b`.

### Potential Issues
1. **Incorrect Handling of Large `a` and Small `b`:**
   - The code might not handle cases where `a` is large and `b` is small correctly. Specifically, the division and ceiling operations might not always produce the correct number of operations.
2. **Edge Cases:**
   - Test cases where `a` is a multiple of `b` or very close to a multiple of `b` can be tricky.

### Challenging Test Input
To create a small-scale test input that exposes the bug, we can use the following cases:
- A case where `a` is a multiple of `b` and `a` is large.
- A case where `a` is slightly more than a multiple of `b`.
- A case where `a` is slightly less than a multiple of `b`.

### Test Input
```plaintext
3
100 2
100 3
100 4
```

### Explanation
1. **Case 1: `100 2`**
   - Here, `a` is 100 and `b` is 2. The code should correctly handle this case, but it's a good starting point to see if the division and ceiling operations are correct.
2. **Case 2: `100 3`**
   - Here, `a` is 100 and `b` is 3. This case is more challenging because 100 is not a multiple of 3, and the division and ceiling operations need to be precise.
3. **Case 3: `100 4`**
   - Here, `a` is 100 and `b` is 4. This case is also challenging because 100 is a multiple of 4, and the code should handle it correctly.

These test cases should help expose any issues in the buggy code, particularly in how it handles large `a` values and different `b` values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
# python cf.py < cf.txt
# https://stackoverflow.com/questions/59060237/how-to-setup-visual-studio-code-stdin-stdout-redirection-for-python-debugger
# https://stackoverflow.com/a/48650929
# запускай отладку alt+q enter

# одно число
# num: int = int(input())

# числа в одной строке
# string: list[str] = open(0).read().split()
# n, k = map(int, string)
# nums: list[int] = list(map(int, strings[str_pos].split()))

# строки
strings: list[str] = open(0).read().split('\n')

# одна строка
# string: str = open(0).read().split('\n')[0]

##################### imports

# from collections import Counter
import math
################################ РЕШЕНИЕ ####################

test_count: int = int(strings[0])

str_pos: int = 1

result: list[str] = []

for i in range(test_count):

    # nums: list[int] = list(map(int, strings[str_pos].split()))
    # nums: str = strings[str_pos]
    n, k = map(int, strings[str_pos].split())

    if k == 2:
        r: int = n - 1
    elif n == 1:
        r: int = 0
    elif k >= n:
        r: int = 1
    else:
        r: int = n // (k - 1)

    result.append(str(r))
    result.append('\n')

    str_pos += 1

    # if one_count > zero_count:
    #     result.append("YES\n")
    # else:
    #     result.append("NO\n")

    # str_pos += 1

################################ ВЫВОД ####################

# print(' '.join(map(str, A)))

# список строк
# print(*result)
print(''.join(result))

# одно число
# print(result)

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios where the logic in the code might fail. The buggy code has specific conditions for different values of \( k \) and \( n \), so we should focus on these:

1. **Edge Case: \( k = 2 \)**:
   - The code assumes that if \( k = 2 \), the number of operations is \( n - 1 \). This is correct, but we should test it with various values of \( n \).

2. **Edge Case: \( n = 1 \)**:
   - The code correctly returns 0 for \( n = 1 \). We should include this case to ensure it works.

3. **Edge Case: \( k \geq n \)**:
   - The code assumes that if \( k \geq n \), the number of operations is 1. This is correct, but we should test it with various values of \( n \) and \( k \).

4. **General Case: \( k < n \)**:
   - The code uses \( r = n // (k - 1) \). This might not always be correct, especially for values of \( k \) and \( n \) where the division does not yield the correct result. We should test it with values that are not straightforward.

5. **Large Values**:
   - Testing with large values of \( n \) and \( k \) can help ensure the code handles them correctly.

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

```plaintext
5
1 5
5 2
6 3
16 4
1000 1000
```

Explanation:
- **Test Case 1**: \( n = 1 \), \( k = 5 \) (Edge case: \( n = 1 \))
- **Test Case 2**: \( n = 5 \), \( k = 2 \) (Edge case: \( k = 2 \))
- **Test Case 3**: \( n = 6 \), \( k = 3 \) (General case: \( k < n \))
- **Test Case 4**: \( n = 16 \), \( k = 4 \) (General case: \( k < n \))
- **Test Case 5**: \( n = 1000 \), \( k = 1000 \) (Edge case: \( k \geq n \))

This test input should help expose any bugs in the given code by covering a variety of edge cases and general scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

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

public class CakeCutting {

    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
        public FastReader()
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        // to read string
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        // to read int
        int nextInt()
        {
            return Integer.parseInt(next());
        }
        // to read long
        long nextLong()
        {
            return Long.parseLong(next());
        }
        // to read double
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
        // // to read float
        float nextFloat()
        {
            return Float.parseFloat(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static void main(String[] args) {
        FastReader fr= new FastReader();
        int T= fr.nextInt();
        StringBuilder s = new StringBuilder();

        for(int t=1;t<=T;t++){

            int n = fr.nextInt();
            int k = fr.nextInt();
            int ans = 0;
            while(n > 1){
                if(n % k == 0){
                    ans += 1 + (n/k);
                    n = 0;
                }
                else {
                   if (n % k == 1) {
                       ans += 1;
                       n -= 1;
                   }
                   else {
                       ans += 2;
                       n -= (n % k);
                   }
                }
            }
            s.append(ans);
            if(t<T)s.append("\n");
        }
        System.out.println(s);
    }
    private static int minOperations(int n, int k) {
        if (n == 1) {
            return 0;
        }

        // Queue to store pairs of current number and number of operations
        Queue<int[]> queue = new LinkedList<>();
        // Set to store visited numbers to avoid reprocessing
        Set<Integer> visited = new HashSet<>();

        queue.offer(new int[]{n, 0});
        visited.add(n);

        while (!queue.isEmpty()) {
            int[] current = queue.poll();
            int currentValue = current[0];
            int operations = current[1];

            // Try to break the currentValue into smaller parts
            for (int i = 1; i <= k; i++) {
                int newValue = currentValue - i;

                if (newValue <= 0) {
                    break;
                }

                if (newValue == 1) {
                    return operations + 1;
                }

                if (!visited.contains(newValue)) {
                    queue.offer(new int[]{newValue, operations + 1});
                    visited.add(newValue);
                }
            }
        }

        return -1; // Should not reach here for valid inputs as per constraints
    }
}

**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 2
2 2
3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
import java.util.Scanner;

public class Cf_P {

    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 ans = 0;

            if(n==1){ans=0;}
            else if(n<=k){ans=1;}

            else{
                if((n/k)<k){
                    if((n%k)==1 || (n%k)==0){ans=1+n/k;}
                    else{ans=1+n/k+1;}
                }
                else if((n/k)==k){
                    if((n%k)==0){ans=1+n/k;}
                    else if((n%k)==1){ans=1+(n/k-1)+2;}
                    else{ans=1+(n/k-1)+1+2;}
                }
                else{
                    int nk = n/k;
                    //n/k>k-->No. of k in n is more than k
                    //nk = n/k + remainder cases
                    ans = nk/(k-1)+nk;
                    if(n%k==1 || n%k==0){ans=ans;}
                    else{ans=ans+1;}
                    }
                }
            System.out.println(ans);
            t--;
        }
        }
    }

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

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

Think step by step.

Response:
```plaintext
4
1 5
15 3
100 10
1024 32
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
// بسم الله
#include <bits/stdc++.h>
#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
#define ll long long
#define true cout << "YES" << '\n'
#define false cout << "NO" << '\n'
#define none cout << "-1" << '\n'
#define fo for (int i = 0; i < n; i++)
using namespace std;
map<pair<char, ll>, long long> mp;
map<long long, long long> mp2;
set<long long> st;
pair<long long, long long> p[210000];
vector<ll> v;
vector<long long> v2;
long long a[210000];
bool s1[210000], b = 0;
ll n, m, d, t, k, cou = 0, mx = 0, x = 10, y = 0, mx1 = 0, cou1 = 0, cou2 = 0, turn = 0;
string s;
void solve()
{
    cin >> n >> k;

    if (n==1)
        cout << 0 << '\n';
    else if (n <= k)
        cout << 1 << '\n';

    else if (n % k == 0)
        cout << 1 + n / k << '\n';
    else
        cout << 2 + n / k << '\n';
}

main()

{
    IOS
            cin >>
        t;
    // t = 1;
    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:
To create a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases or scenarios that the code might handle incorrectly. The provided buggy code has a few potential issues:

1. **Incorrect handling of cases where `n` is not a multiple of `k` but is close to a multiple of `k`**.
2. **Incorrect handling of cases where `n` is a small multiple of `k` plus a small remainder**.

Let's consider a test case where `n` is a small multiple of `k` plus a small remainder. This will test the logic in the `else` block of the code, which is where the bug is most likely to be exposed.

### Test Case Analysis
- **Case 1: `n = 7` and `k = 3`**
  - The correct number of operations should be 3.
  - The buggy code will incorrectly output 4 because it uses the formula `2 + n / k`, which is not always correct.

### Test Input
```plaintext
1
7 3
```

### Explanation
- **Input: `1`** indicates that there is one test case.
- **Input: `7 3`** sets `n = 7` and `k = 3`.

### Expected Output
- The correct output should be `3`.

### Buggy Code Output
- The buggy code will output `4`.

This test case will expose the bug in the logic where the code incorrectly handles the case where `n` is not a multiple of `k` but is close to a multiple of `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

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

using namespace std;

typedef long long ll;
typedef pair<int, int>PII;
const int N=2e5+10;
const int MOD = 1e9 + 7;
const int INF=0X3F3F3F3F;
const int dx[]={-1,1,0,0,-1,-1,+1,+1};
const int dy[]={0,0,-1,1,-1,+1,-1,+1};
const int M = 1e7 + 10;

int t;

int main()
{
	cin >> t;
	while(t --)
	{
		int n, k;
		cin >> n >> k;
		if(n <= k) cout << 0 << endl;
		else {
			int cnt = 1, s = 0;
			cout << (n + k - 1) / k + 1 << 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
5
1 2
2 2
3 2
4 2
5 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
from sys import stdin

def istr():
    return str(stdin.readline().strip())

def lstr():
    return list(map(str, list(stdin.readline().strip())))

def iint():
    return int(stdin.readline().strip())

def lint():
    return list(map(int, stdin.readline().strip().split()))

def bint():
    return list(map(int, list(stdin.readline().strip())))

def sol():
    n, k = lint()
    ans = 0
    extra = 0
    while n >= k or n not in [1, 0]:
        if n % k != 0:
            extra += 1
            n -= n % k
        ans += n // k
        n //= k
    print(ans + extra)

ttt = iint()
# ttt = 1
for __ in range(ttt):
    sol()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

Buggy Code:
def solve():
    n,k = map(int,input().split())
    if n == 1:
        print(0)
        return
    if n%k !=0:
        ans = n//k+2
    else:
        ans = n//k+1
    print(ans)

for __ in range(int(input())):
    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
4
1 5
7 2
8 3
15 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

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

public class Main {
    public static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    public static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));

    public static int nextInt() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }

    //如果数字太大会有精度丢失
    public static long nextLong() throws IOException {
        in.nextToken();
        return (long) in.nval;
    }

    //如果数字太大会有精度丢失
    public static double nextDouble() throws IOException {
        in.nextToken();
        return in.nval;
    }

    //字符串不能以数字开头，字符串不能是特殊符号，空格(| _ % ^..... )
    public static String next() throws IOException {
        in.nextToken();
        return in.sval;
    }
    static long lcm(int a, int b) {
        return (long)a * b / gcd(a,b);
    }
    static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }
    public static long pow(long x, int n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = ans * x % MOD;
            }
            x = x * x % MOD;
            n >>= 1;
        }
        return ans;
    }
    static final int MOD = 998244353;
    static final int N = 500;
    static int dfs(int n,int k){
        if(n == 1){
            return 0;
        }
        if(n <= k) {
             return 1;
        }
        int res  = Integer.MAX_VALUE;
        for(int i = 2;i<=k;i++){
            int t = n / i;
            int mod = n % i;
            if(mod == 0){
                res = Math.min(res,i * dfs(t,k));
            }else{
                res = Math.min(res,(i - 1) * dfs(t,k) + dfs(t + mod ,k));
            }
        }
        return res + 1;
    }
    public static void main(String[] args) throws IOException{
        int t = nextInt();
        while (t-- > 0) {
            int n = nextInt();
            int k = nextInt();
            pw.println(dfs(n,k));
        }

        pw.flush();
    }
}

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

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

Think step by step.

Response:
```plaintext
4
1 5
5 2
6 3
16 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Split the Multiset
time_limit_ms: 1000
memory_limit_mb: 512
Description: A multiset is a set of numbers in which there can be equal elements, and the order of the numbers does not matter. For example, $\\{2,2,4\\}$ is a multiset.

You have a multiset $S$. Initially, the multiset contains only one positive integer $n$. That is, $S=\\{n\\}$. Additionally, there is a given positive integer $k$.

In one operation, you can select any positive integer $u$ in $S$ and remove one copy of $u$ from $S$. Then, insert no more than $k$ positive integers into $S$ so that the sum of all inserted integers is equal to $u$.

Find the minimum number of operations to make $S$ contain $n$ ones.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

The only line of each testcase contains two integers $n,k$ ($1\le n\le 1000,2\le k\le 1000$).

Output: For each testcase, print one integer, which is the required answer.

Examples:
input:
4
1 5
5 2
6 3
16 4
output:
0
4
3
5

Note: For the first test case, initially $S=\\{1\\}$, already satisfying the requirement. Therefore, we need zero operations.

For the second test case, initially $S=\\{5\\}$. We can apply the following operations:

  * Select $u=5$, remove $u$ from $S$, and insert $2,3$ into $S$. Now, $S=\\{2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,2$ into $S$. Now, $S=\\{1,1,1,2\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,1,1\\}$.

Using $4$ operations in total, we achieve the goal.

For the third test case, initially $S=\\{6\\}$. We can apply the following operations:

  * Select $u=6$, remove $u$ from $S$, and insert $1,2,3$ into $S$. Now, $S=\\{1,2,3\\}$.    * Select $u=2$, remove $u$ from $S$, and insert $1,1$ into $S$. Now, $S=\\{1,1,1,3\\}$.    * Select $u=3$, remove $u$ from $S$, and insert $1,1,1$ into $S$. Now, $S=\\{1,1,1,1,1,1\\}$.

Using $3$ operations in total, we achieve the goal.

For the fourth test case, initially $S=\\{16\\}$. We can apply the following operations:

  * Select $u=16$, remove $u$ from $S$, and insert $4,4,4,4$ into $S$. Now, $S=\\{4,4,4,4\\}$.    * Repeat for $4$ times: select $u=4$, remove $u$ from $S$, and insert $1,1,1,1$ into $S$.

Using $5$ operations in total, we achieve the goal.

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

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

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

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

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

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

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

        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		try {
		    FastReader scan = new FastReader();
		    int t = scan.nextInt();

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

		        if (n == 1) {
	                System.out.println(0);
	                continue;
	            }

	            int operations = 0;
	            int current = n;

	            while (current > 1) {
	                operations++;
	                if (current <= k) {
	                    break;
	                }
	                int parts = Math.min(k, current);
	                current = (current + parts - 1) / parts;
	            }

	            operations += current - 1;
	            System.out.println(operations);
		    }

		} catch(Exception e) {
		    return;
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
5
1 5
5 2
6 3
16 4
1000 1000
```