**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
import java.util.*;

public class Solution {
    private static final char NL = '\n';
    private static final int INF = 1_000_000_000;

    public static void main(String[] args) {
        Scanner io = new Scanner(System.in);
        int num = io.nextInt();
        for (int i = 0; i < num; i++) {
            solve(io);
        }
        io.close();
    }
    public static void solve(Scanner io) {
        List<Integer> a = new ArrayList<>();
        {
            int n;
            n = io.nextInt();

            Map<Integer, Integer> cnt = new HashMap<>();
            while (n-- > 0) {
                int x = io.nextInt();
                cnt.put(x, cnt.getOrDefault(x, 0) + 1);
            }

            for (Map.Entry<Integer, Integer> entry : cnt.entrySet()) {
                a.add(entry.getValue());
            }
        }

        int n = a.size();
        int[] dp = new int[n + 1];
        Arrays.fill(dp, INF);
        dp[0] = 0;

        for (int i = 1; i <= n; i++) {
            int[] ndp = dp.clone();

            for (int k = 1; k <= n; k++) {
                int nv = dp[k - 1] + a.get(i - 1);
                if (nv <= i - k) {
                    ndp[k] = Math.min(ndp[k], nv);
                }
            }

            dp = ndp;
        }

        int ans = n;
        while (dp[ans] >= INF) ans--;
        System.out.println(n - ans + NL-10);
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define time cerr << "\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s.\n" ;
#define tsree3 ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std ;
int main()
{
    //freopen("input.txt","r",stdin) ;
    //freopen("output.txt","w",stdout) ;
    tsree3
    ll t = 1 ;
    cin >> t ;
    while(t--)
    {
        ll n = 0 , m = 0 , k = 0 , num = 0 ;
        cin >> n >> m >> k ;
        if(n>k&&m>k)
        {
            cout << k*k << "\n" ;
        }
        else if(n<k&&m<k)
        {
            cout << n*m << "\n" ;
        }
        else if(n<=k)
        {
            cout << n*k << "\n" ;
        }
        else if(m<=k)
        {
            cout << m*k << "\n" ;
        }
        else if(n==m&&m==k)
        {
            cout << n*m << "\n" ;
        }
    }
    time
    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
3 3 2
4 4 2
5 5 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
#include<stdio.h>
#include<algorithm>
int  main()
{
	int t;
	scanf("%d",&t);
	while(t--){
		int n,m,k;
		int c=0;
	scanf("%d %d %d",&n,&m,&k);
	if(n>k&&m>k) c=k*k;
	if(n<k&&m<k) c=n*m;
	if(n>k&&m<=k)  c=m*k;
	if(n<=k&&m>k)  c=n*k;
	printf("%d\n",c);
	}

	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
#include<iostream>
#include<cstdio>
#include<math.h>
#include<string>
#include<cmath>
#include<ctype.h>
#include<vector>
#include<algorithm>
#include<utility>
#include<list>
#include<set>
#include<map>
#include<stack>
#include<queue>
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define vl vector<long long>
#define mapl map<long long ,long long>
#define ll long long
///#include<windows.h>
///(n+1)/2*n
///#define int long long
///#include<bits/stdc++.h>
using namespace std;
int main()
{
   fast
   int t;
   cin>>t;
   while(t--){
    ll n,m,k;
    cin>>n>>m>>k;
    //k--;
    if(k==1)cout<<1<<endl;
    else if(k>n*m || (n<k and m<k))cout<<n*m<<endl;
    else{
        if(n<=k)cout<<k*n<<endl;
        else if(m<=k)cout<<m*(n-(n-k))<<endl;
        else cout<<k*(max(n,m)-(max(n,m)-k))<<endl;
            ////cout<<(k-1)-abs(min(n,m)-(k-1))<<endl;
    }
   }
   return 0;
}
///thank god for every thing and anything.....

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

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

Think step by step.

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

### Analysis of the Buggy Code
1. **Case 1: `k == 1`**
   - The code outputs `1` if `k == 1`.
   - This is correct because if `k == 1`, the minimum number of colors needed is `1`.

2. **Case 2: `k > n * m` or `(n < k and m < k)`**
   - The code outputs `n * m`.
   - This is correct because if `k` is greater than the dimensions of the grid, each cell must have a unique color.

3. **Case 3: `n <= k`**
   - The code outputs `k * n`.
   - This is incorrect. If `n <= k`, the number of colors needed is `m` because each row can be colored independently.

4. **Case 4: `m <= k`**
   - The code outputs `m * (n - (n - k))`, which simplifies to `m * k`.
   - This is incorrect. If `m <= k`, the number of colors needed is `n` because each column can be colored independently.

5. **Case 5: `k <= n` and `k <= m`**
   - The code outputs `k * (max(n, m) - (max(n, m) - k))`, which simplifies to `k * k`.
   - This is incorrect. The correct number of colors needed is `k * k` only if both `n` and `m` are at least `k`.

### Challenging Test Input
To expose the bugs, we need to create test cases that fall into the incorrect branches of the code.

1. **Test Case 1: `n <= k` and `m > k`**
   - This should expose the bug in the `n <= k` case.
   - Example: `n = 3`, `m = 5`, `k = 3`.

2. **Test Case 2: `m <= k` and `n > k`**
   - This should expose the bug in the `m <= k` case.
   - Example: `n = 5`, `m = 3`, `k = 3`.

3. **Test Case 3: `k <= n` and `k <= m`**
   - This should expose the bug in the `k <= n` and `k <= m` case.
   - Example: `n = 5`, `m = 5`, `k = 3`.

### Output
```plaintext
3
3 5 3
5 3 3
5 5 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
num_test_cases = int(input())
test_cases = []

for _ in range(num_test_cases):

    n, m, k = map(int, input().split())
    test_cases.append((n, m, k))
for i in range(num_test_cases):
    if(test_cases[i][2]< min(test_cases[i][1],test_cases[i][0])):
        q= max(test_cases[i][1],test_cases[i][0])
        print(2*(test_cases[i][2])*(q-test_cases[i][2]))
    elif(test_cases[i][2]> max(test_cases[i][1],test_cases[i][0])):
        print(test_cases[i][0]*test_cases[i][1])
    elif(test_cases[i][2]== max(test_cases[i][1],test_cases[i][0])):
        print(test_cases[i][1]*test_cases[i][0])
    elif(test_cases[i][2]== min(test_cases[i][1],test_cases[i][0])):
        s=min(test_cases[i][0],test_cases[i][1])
        print(s**2)

    else:
        z= min(test_cases[i][1],test_cases[i][0])
        print(test_cases[i][2]*z)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the key points to consider:

1. **Small Grid with Large \( k \)**: When \( k \) is larger than both \( n \) and \( m \), the code incorrectly assumes that the number of colors needed is \( n \times m \). This is a common edge case where the grid is too small to have any cells with the same color.

2. **Equal \( n \) and \( m \)**: When \( n \) and \( m \) are equal and \( k \) is also equal to \( n \), the code incorrectly prints \( s^2 \). This is another edge case where the grid is a square and \( k \) is equal to the side length.

3. **\( k \) is Between \( n \) and \( m \)**: When \( k \) is between \( n \) and \( m \), the code incorrectly prints \( 2 \times k \times (q - k) \). This is a scenario where the grid dimensions and \( k \) are not aligned in a straightforward way.

4. **Small \( k \)**: When \( k \) is small compared to \( n \) and \( m \), the code might not correctly calculate the number of colors needed.

Based on these points, let's create a small-scale test input that includes these edge cases:

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

Explanation:
- **Case 1 (1 1 100)**: \( k \) is much larger than both \( n \) and \( m \). The correct output should be 1.
- **Case 2 (2 2 2)**: \( n \), \( m \), and \( k \) are all equal. The correct output should be 4.
- **Case 3 (3 3 3)**: \( n \), \( m \), and \( k \) are all equal. The correct output should be 9.
- **Case 4 (4 4 2)**: \( k \) is smaller than \( n \) and \( m \). The correct output should be 4.
- **Case 5 (10 10 5)**: \( k \) is between \( n \) and \( m \). The correct output should be 25.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
t= int(input().strip())

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

    if k>m and k>n:
        print(n*m)
        continue

    colour = 0
    color_mapping = set()

    for i in range(n):
        for j in range(m):
            if (i, j) in color_mapping:
                continue
            colour += 1
            color_mapping.add((i, j))
            color_mapping.add((i, j + k))
            color_mapping.add((i + k, j))
            color_mapping.add((i + k, j + k))

    print(colour)

**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 1
3 3 2
4 4 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
def getResult(n,m,k):
    if n*m <= k:
        return n*m

    if m>k and n>k:
        return k**2
    else:
        if m<k:
            return m*k
        else:
            return n*k

t = int(input())
for i in range(t):
    n,m,k = input().split()
    n = int(n)
    m = int(m)
    k = int(k)

    print(getResult(n,m,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
3
10 10 3
3 3 3
100 100 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
//     package CP;

import java.io.IOException;
import java.util.*;
public class cf {
    //     static BufferedReader sd = new BufferedReader(new InputStreamReader(System.in));
    static Scanner sd = new Scanner(System.in);

    public static boolean IsPrime(int x) {
        for (int i = 2; i * i <= x; i++) {
            if (x % i == 0)
                return false;
        }
        return true;
    }

    public static int gcd(int a, int b) {
        if (a % b == 0)
            return b;
        return gcd(b, a % b);
    }

    public static void input(long[] arr) {
        for (int i = 0; i < arr.length; i++) arr[i] = sd.nextLong();
    }

    static int mod = 1000000000 + 7;

    public static void main(String[] args) throws IOException {
        int t = sd.nextInt();
        while (t-- > 0) {
            run();
        }
    }

    /*****************************************CODE START HERE*********************************************************/
    public static void run()
    {
        int n = sd.nextInt();
        int m = sd.nextInt();
        int k = sd.nextInt();
        if (k>=(n*m))
        {
            System.out.println(n*m);
        }
        else
        {
            int a = (n*m)/2;
            int ans =0;
            int p =1;
            while (ans<a)
            {
                ans =k*p;
                p++;
            }
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
import java.util.*;

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

        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();
            // System.out.println(n);
            // System.out.println(m);
            // System.out.println(k);
            if(n*m<k){
                System.out.println(n*m);
            }
            else
             if(n>k && m<k)
            {
                System.out.println(m*k);
            }
            else
            if(n<k && m>k)
            {
                System.out.println(n*k);
            }
            else if(m>k && n>k){
                System.out.println(k*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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has several conditions that might not handle certain cases correctly. Let's analyze the conditions and create a test input that can expose the bug.

### Analysis of the Buggy Code
1. **If `n * m < k`**: This condition checks if the total number of cells is less than the magic constant `k`. If true, it prints `n * m`.
2. **If `n > k` and `m < k`**: This condition checks if the number of rows is greater than `k` and the number of columns is less than `k`. If true, it prints `m * k`.
3. **If `n < k` and `m > k`**: This condition checks if the number of rows is less than `k` and the number of columns is greater than `k`. If true, it prints `n * k`.
4. **If `m > k` and `n > k`**: This condition checks if both the number of rows and columns are greater than `k`. If true, it prints `k * k`.

### Edge Cases to Consider
1. **When `n * m < k`**: This is a straightforward case where the grid is too small to have any cells with the same color.
2. **When `n > k` and `m < k`**: This case should be handled correctly, but we need to check if the logic is sound.
3. **When `n < k` and `m > k`**: Similar to the previous case, we need to ensure the logic is correct.
4. **When `m > k` and `n > k`**: This is a common case where the grid is larger than `k` in both dimensions.
5. **When `n == k` or `m == k`**: These are edge cases where the grid size exactly matches the magic constant `k`.

### Challenging Test Input
Let's create a test input that includes these edge cases:

```plaintext
5
3 3 10
10 3 10
3 10 10
10 10 10
5 5 5
```

### Explanation
1. **3 3 10**: Here, `n * m < k` should be true, so the output should be `9`.
2. **10 3 10**: Here, `n > k` and `m < k` should be true, so the output should be `30`.
3. **3 10 10**: Here, `n < k` and `m > k` should be true, so the output should be `30`.
4. **10 10 10**: Here, `m > k` and `n > k` should be true, so the output should be `100`.
5. **5 5 5**: This is an edge case where `n == k` and `m == k`. The output should be `25`.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

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

using namespace std;

int main(){

    int t;
    cin >> t;
    while(t--){
        int n,m,k;
        cin >> n >> m >> k;
        if(n * m <= k) {
            cout << n*m << endl;
        }
        else {
            if((n * m) / (k * k) > 1) {
                cout << k * k << endl;
            }
            else if((n * m) / (k * k) == 1) {
                cout << (k*k) - k << endl;
            }
            else {
                cout << (k*k) - 2*k << endl;
            }
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3 3 2
5 5 2
8 8 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
#include<iostream>
using namespace std;
int main()
{
    int t;
    cin>>t;
    int ans=0;
    for(int i=0;i<t;i++)
    {
        int m,n;
        cin>>m>>n;
        int k;
        cin>>k;
        int l=max(m,n);
        int s=min(m,n);
      if(k<l && k<s)
      {
        ans=(k)*k;
      }
      if(k<l && k>=s)
      {
        ans=k*s;
      }
      if(k>l)
      {
        ans=m*n;
      }
       cout<<ans<<endl;
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the logic in the code might fail. The buggy code has the following issues:

1. It does not correctly handle the case where `k` is greater than both `m` and `n`.
2. It does not correctly handle the case where `k` is greater than or equal to one dimension but less than the other.
3. It does not correctly handle the case where `k` is less than both dimensions but the grid size is large.

Let's consider these scenarios and create a small-scale test input:

1. **Case where `k` is greater than both `m` and `n`**:
   - This should result in `m * n` colors.
   - Example: `m = 3`, `n = 3`, `k = 5`.

2. **Case where `k` is greater than one dimension but less than the other**:
   - This should result in `m * n` colors.
   - Example: `m = 5`, `n = 3`, `k = 4`.

3. **Case where `k` is less than both dimensions but the grid size is large**:
   - This should result in a more complex calculation.
   - Example: `m = 6`, `n = 6`, `k = 2`.

4. **Case where `k` is equal to one dimension**:
   - This should result in a specific number of colors.
   - Example: `m = 4`, `n = 4`, `k = 4`.

5. **Case where `k` is very small compared to both dimensions**:
   - This should result in a specific number of colors.
   - Example: `m = 10`, `n = 10`, `k = 1`.

Here is the test input:

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

This input covers various edge cases and should expose the bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
for _ in range(int(input())):
     n, m, k=map(int, input().split())
     if n*m<k:
          print(n*m)
     else:
          print(min(min(n*k, m*k), min(k*k, m*k)))

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

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

Think step by step.

Response:
```plaintext
1
10 10 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
n= int(input())
l=[]
for i in range (0,n):
    l1= list(map(int,input().split()))
    l.append(l1)
for k in l:
    m=k[0]
    n=k[1]
    l=k[2]
    k.sort()
    if(k[1]==l):
        print(k[1]*k[0])
    if(k[0]==l):
        print(k[0]*k[0])
    if(k[2]==l):
        print(k[0]*k[1])

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify its weaknesses. The buggy code has the following issues:

1. It incorrectly assigns the values of `m`, `n`, and `l` to the elements of the list `k`.
2. It sorts the list `k` which can lead to incorrect assignments of `m`, `n`, and `l`.
3. It has incorrect logic for determining the number of colors needed.

To expose these issues, we can create test cases that:
- Test the sorting issue by providing inputs where `m`, `n`, and `l` are not in ascending order.
- Test the logic for determining the number of colors needed by providing inputs where `k` is equal to `m` or `n` and where `k` is less than both `m` and `n`.

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

```plaintext
4
3 3 2
5 1 10000
7 3 4
3 2 7
```

Explanation:
1. **3 3 2**: This is a standard case where `k` is less than both `m` and `n`. The correct output should be 4.
2. **5 1 10000**: This is a case where `k` is much larger than both `m` and `n`. The correct output should be 5.
3. **7 3 4**: This is a case where `k` is less than one dimension but greater than the other. The correct output should be 12.
4. **3 2 7**: This is a case where `k` is greater than both `m` and `n`. The correct output should be 6.

These test cases should help identify the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
import java.util.*;

public class distColor {

    public static void main(String args[]){

        try(Scanner sc = new Scanner(System.in)){

            int test = sc.nextInt();

            while((test--)>0){

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

                int p = (n/k);
                int p1 = (n%k);
                int q = (m/k);
                int q1 = (m%k);
                int ans = 0;
                // if(n>=m){
                //     ans = (p+p1)*(q+q1);
                // }else{
                //     ans = (p*k)*(q*k);
                // }
                if(p==0 && q==0){
                    ans = p1*q1;
                }else if(p!=0 && q==0){
                    ans = (p*k)*q1;
                }else if(p==0 && q!=0){
                    ans = (p1)*(q*k);
                }else{
                    ans = (p*k)*(q*k);
                }
                // ans = (p*k)*(q*k);

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

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues.

### Analysis of the Buggy Code
1. **Division and Modulo Operations**:
   - The code divides `n` and `m` by `k` to get `p` and `q` respectively.
   - It also calculates the remainders `p1` and `q1`.

2. **Conditions**:
   - The code has several conditions to determine the answer, but it seems to be missing the correct logic to handle the grid coloring problem correctly.
   - The code seems to assume that the answer is always the product of `p*k` and `q*k` in most cases, which is incorrect.

### Edge Cases to Consider
1. **Small Grids**:
   - When `n` and `m` are small and `k` is large, the grid might not be divisible by `k` and the remainders play a significant role.
2. **Equal Dimensions**:
   - When `n` and `m` are equal, the grid is a square, and the logic should handle this case correctly.
3. **Large `k`**:
   - When `k` is very large compared to `n` and `m`, the grid might be completely or almost completely covered by the remainders.
4. **Non-integer Division**:
   - When `n` and `m` are not perfectly divisible by `k`, the remainders should be considered in the calculation.

### Challenging Test Input
To expose the bug, we can create a test case where the remainders `p1` and `q1` are non-zero and play a significant role in the answer. Here is a small-scale test input that fits the criteria:

```plaintext
4
3 3 2
5 1 10000
7 3 4
3 2 7
```

### Explanation
1. **3 3 2**:
   - `n = 3`, `m = 3`, `k = 2`
   - `p = 1`, `p1 = 1`
   - `q = 1`, `q1 = 1`
   - The correct answer is 4, but the buggy code might incorrectly calculate it.

2. **5 1 10000**:
   - `n = 5`, `m = 1`, `k = 10000`
   - `p = 0`, `p1 = 5`
   - `q = 0`, `q1 = 1`
   - The correct answer is 5, but the buggy code might incorrectly calculate it.

3. **7 3 4**:
   - `n = 7`, `m = 3`, `k = 4`
   - `p = 1`, `p1 = 3`
   - `q = 0`, `q1 = 3`
   - The correct answer is 12, but the buggy code might incorrectly calculate it.

4. **3 2 7**:
   - `n = 3`, `m = 2`, `k = 7`
   - `p = 0`, `p1 = 3`
   - `q = 0`, `q1 = 2`
   - The correct answer is 6, but the buggy code might incorrectly calculate it.

This test input should help expose the bugs in the provided code by focusing on edge cases where the remainders play a significant role.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
import java.util.*;

public class A_Distanced_Coloring {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int testCases = sc.nextInt();
        while (testCases-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();
            int[][] mat = new int[n][m];
            int count = 0;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (mat[i][j] == 0) {
                        count++;
                        mat[i][j] = count;
                        if (i + k < n) {
                            mat[i + k][j] = count;
                        }
                        if (j + k < m) {
                            mat[i][j + k] = count;
                        }
                        if (i + k < n && j + k < m) {
                            mat[i + k][j + k] = count;
                        }
                    }
                }
            }
            System.out.println(count);
        }
        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
#include <bits/stdc++.h>
// #include <iostream>
// #include <vector>
// #include <algorithm>
// #include <string>
// #include <cstdlib>
// #include <ctime>

using namespace std;

int main(){
    int num;
    cin >> num;

    vector<int> n(num), m(num), k(num);

    for(int i = 0; i < num; i++){
        cin >> n[i] >> m[i] >> k[i];
    }

    vector<int> minColor(num, 0);

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

        vector<vector<int> > visited(n[i], vector<int>(m[i], 0));
        bool allCovered = false;
        int x_inc=0, y_inc=0;
        while (!allCovered) {
            minColor[i]++;

            for(int q = x_inc; q < n[i]; q += k[i]) {
                for(int r = y_inc; r < m[i]; r += k[i]) {
                    visited[q][r]++;
                }
            }

            allCovered = true;
            for(int q = 0; q < n[i]; q++) {
                for(int r = 0; r < m[i]; r++) {
                    if(visited[q][r] == 0) {
                        allCovered = false;
                        x_inc =q;
                        y_inc= r;
                        break;
                    }
                }
                if(!allCovered) break;
            }
        }
    }

    for(int i = 0; i < num; i++){
        cout << minColor[i] << endl;
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
#include <bits/stdc++.h>
#include <iostream>
#include <algorithm>
#define ll long long
#define all(x) (x).begin(), (x).end()
#define fr(n) for(ll i = 0 ; i < n ; i++)
#define frj(n) for(ll j = 0 ; j < n ; j++)
#define endl '\n'
#define FAST cin.tie(0), cout.tie(0), cin.sync_with_stdio(0), cout.sync_with_stdio(0);
using namespace std;
const int MOD = 1e9 + 7;
const ll N = 2e6 + 5;
int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, -1, 0, 1, -1, 1, -1, 1};
char di[] = {'D', 'L', 'U', 'R'};
ll n,m,k;
ll gcd(ll a, ll b) {
    if(b == 0) return a;
    return gcd(b, a % b);
}

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

ll add(ll a, ll b) {
    return (a % MOD + b % MOD) % MOD;
}

bool prime[N + 1];
vector<ll> primes;

// Function to determine primes in range
void sieve() {
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1] = false; // 0 and 1 are not primes
    for (int i = 2; i <= N; i++) {
        if (prime[i]) {
            primes.push_back(i);
            for (int j = i * 2; j <= N; j += i) {
                prime[j] = false;
            }
        }
    }
}

// Smallest prime factor array
int SPF[N + 1];
void sieveSPF() {
    for (int i = 0; i <= N; i++)
        SPF[i] = i;

    for (int i = 2; i * i <= N; i++) {
        if (SPF[i] == i) { // i is a prime
            for (int j = i * i; j <= N; j += i) {
                if (SPF[j] == j) { // j is not marked yet
                    SPF[j] = i;
                }
            }
        }
    }
}

map<ll, int> factorization;

void factorize(ll x){
    for (auto p : primes) {
        if (p * p > x)
            break;

        while (x % p == 0) {
            factorization[p]++;
            x /= p;
        }
    }
    if (x != 1)
        factorization[x]++;
}

vector<pair<ll, int>> vfactorize;
vector<int> divisors;

void getDivisors(int ind, ll div) {
    if (ind == vfactorize.size()) {
        divisors.push_back(div);
        return;
    }
    for (int i = 0; i <= vfactorize[ind].second; i++) {
        getDivisors(ind + 1, div);
        div *= vfactorize[ind].first;
    }
}

vector<int>adj[10000];
bool vis[10000];
int an=0;
set<int>ans;
void dfs(int node)
{
    vis[node]=true;
    for(auto  it : adj[node])
    {
        if(vis[it])
        {
            if(ans.size()==n/2)return;
            ans.insert(it);
        }else{
            vis[it]=true;
            dfs(it);
        }
    }

}

int orientation(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3) {
    ll val = (y2 - y1) * (x3 - x2) - (x2 - x1) * (y3 - y2);
    if (val == 0) return 0; // collinear
    return (val > 0) ? 1 : 2; // clock or counterclock wise
}

// Function to check if point q lies on line segment pr
bool onSegment(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3) {
    if (x2 <= max(x1, x3) && x2 >= min(x1, x3) &&
        y2 <= max(y1, y3) && y2 >= min(y1, y3))
       return true;
    return false;
}

// The main function that returns true if the line segment 'p1q1' and 'p2q2' intersect.
bool doIntersect(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3, ll x4, ll y4) {
    // Find the four orientations needed for general and special cases
    int o1 = orientation(x1, y1, x2, y2, x3, y3);
    int o2 = orientation(x1, y1, x2, y2, x4, y4);
    int o3 = orientation(x3, y3, x4, y4, x1, y1);
    int o4 = orientation(x3, y3, x4, y4, x2, y2);

    // General case
    if (o1 != o2 && o3 != o4)
        return true;

    // Special Cases
    // p1, q1 and p2 are collinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(x1, y1, x3, y3, x2, y2)) return true;

    // p1, q1 and q2 are collinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(x1, y1, x4, y4, x2, y2)) return true;

    // p2, q2 and p1 are collinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(x3, y3, x1, y1, x4, y4)) return true;

    // p2, q2 and q1 are collinear and q1 lies on segment p2q2
    if (o4 == 0 && onSegment(x3, y3, x2, y2, x4, y4)) return true;

    return false; // Doesn't fall in any of the above cases
}
bool isSubsequence(const string& s, const string& t) {
    int m = s.size(), n = t.size();
    int j = 0;

    for (int i = 0; i < m && j < n; i++) {
        if (s[i] == t[j]) {
            j++;
        }
    }
    return (j == n);
}

void solve()
{
    cin >> n >> m >> k;
        if (n < k && m < k) {
            cout<<n*m;
        } else if (n < k && m >= k) {
            cout<< n*k;
        } else if (n >= k && m < k) {
            cout<<m*k;
        } else {
            cout<<pow(k,2);
        }

        cout <<endl;

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

    ll t=1;
    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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code provided has some logical issues, particularly in how it handles the conditions for determining the minimum number of colors needed.

### Analyzing the Buggy Code:
1. **Case 1: `n < k && m < k`**:
   - The code outputs `n * m`. This is incorrect because if both `n` and `m` are less than `k`, it means each cell must have a different color, and the number of colors required is `n * m`.
   - This case is correct and should not be a source of bugs.

2. **Case 2: `n < k && m >= k`**:
   - The code outputs `n * k`. This is incorrect because the number of colors required should be `m` since each row can be colored with a different set of `k` colors.
   - This case is incorrect and should be a source of bugs.

3. **Case 3: `n >= k && m < k`**:
   - The code outputs `m * k`. This is incorrect because the number of colors required should be `n` since each column can be colored with a different set of `k` colors.
   - This case is incorrect and should be a source of bugs.

4. **Case 4: `n >= k && m >= k`**:
   - The code outputs `k^2`. This is incorrect because the number of colors required is not always `k^2`. It should be the minimum of `n` and `m` times `k`.
   - This case is incorrect and should be a source of bugs.

### Challenging Test Input:
To expose the bugs, we need to create test cases that fall into the incorrect conditions:

1. **Test Case 1: `n < k && m >= k`**:
   - Example: `n = 2`, `m = 4`, `k = 3`
   - Expected output: `4` (since each row can be colored with 4 colors, and each row is independent).

2. **Test Case 2: `n >= k && m < k`**:
   - Example: `n = 4`, `m = 2`, `k = 3`
   - Expected output: `4` (since each column can be colored with 4 colors, and each column is independent).

3. **Test Case 3: `n >= k && m >= k`**:
   - Example: `n = 5`, `m = 5`, `k = 3`
   - Expected output: `9` (since the minimum of `n` and `m` is 5, and each block of 3x3 requires 9 colors).

### Test Input:
```plaintext
3
2 4 3
4 2 3
5 5 3
```

This test input will help expose the bugs in the provided code by covering the incorrect conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
def solve(n,m,k):
	x,y = 0,0
	num_of_colors_used = 1
	same_color_set = []
	for i in range(n):
		for j in range(m):
			if max(abs(x -i),abs(y - j)) >= k:
				same_color_set.append({i,j})
	return ((n*m) - (len(same_color_set)+1)) + num_of_colors_used

t = int(input())
for _ in range(t):
	n,m,k = map(int,input().split())
	ans = solve(n,m,k)
	print(ans)

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

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

Think step by step.

Response:
```plaintext
1
10 10 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
def fill_all_blocks_at_distance_k(arr, n, m, k, i, j,min):
    for r in range(n):
        for c in range(m):
            if (max(abs(r-i),abs(c-j)) >= k):
                arr[r][c] = min

def solve():
    t = int(input())
    for _ in range(t):
        n, m, k = map(int, input().split())

        arr = [[0 for _ in range(m)] for _ in range(n)]
        # print(arr)

        # m = {}
        min = 1

        for i in range(n):
            for j in range(m):
                if(arr[i][j] == 0):
                    arr[i][j] = min
                    # m[min] = (i, j)
                    fill_all_blocks_at_distance_k(arr, n, m, k, i, j, min)
                    min += 1

        print(min-1)

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
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
import java.util.*;

public class ProblemA {
    public static void helper(int i, int j, int n, int m, int[][] grid, int c, int k)
    {
        for(int x = i; x < n; x++)
        {
            for(int y = j; y<m; y++)
            {
                if(Math.max(Math.abs(x - i), Math.abs(y - j)) >= k)
                {
                    grid[x][y] = c;
                }
            }
        }
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();

            int[][] grid = new int[n][m];
            int c = 1;
            for(int i = 0; i<n; i++)
            {
                for(int j = 0; j<m; j++)
                {
                    if(grid[i][j] == 0)
                    {
                        grid[i][j] = c;
                        helper(i, j, n, m, grid, c, k);
                        c++;
                    }

                }
            }

//            for(int i = 0; i<n; i++)
//            {
//                for(int j = 0; j<n; j++)
//                {
//                    System.out.print(grid[i][j] + " ");
//                }
//                System.out.println();
//            }

            System.out.println(c-1);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distanced Coloring
time_limit_ms: 1000
memory_limit_mb: 256
Description: You received an $n\times m$ grid from a mysterious source. The source also gave you a magic positive integer constant $k$.

The source told you to color the grid with some colors, satisfying the following condition:

  * If $(x_1,y_1)$, $(x_2,y_2)$ are two distinct cells with the same color, then $\max(|x_1-x_2|,|y_1-y_2|)\ge k$.

You don't like using too many colors. Please find the minimum number of colors needed to color the grid.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le1000$). The description of the test cases follows.

The only line of each test case consists of three positive integers $n$, $m$, $k$ ($1\le n,m,k\le10^4$) — the dimensions of the grid and the magic constant.

Output: For each test case, print a single integer — the minimum number of colors needed to color the grid.

Examples:
input:
6
3 3 2
5 1 10000
7 3 4
3 2 7
8 9 6
2 5 4
output:
4
5
12
6
36
8

Note: In the first test case, one of the optimal constructions is:

![](CDN_BASE_URL/a1b4551a26d9369b34abeb1ee2f829ed)

In the second test case, the color of all cells must be pairwise different, so the answer is $5$.

Buggy Code:
import java.util.Scanner;

public class A_Distanced_Coloring {
    public static void applyColor(int[][] colors, int color, int i, int j, int n, int m, int k){
        for(int a = i; a < n; a += k){
            for(int b = j; b < m; b += k){
                colors[a][b] = color;
            }
        }
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();
            int[][] colors = new int[n][m];
            int color = 1;
            for(int i = 0; i < n; i++){
                for(int j = 0; j < m; j++){
                    if(colors[i][j] == 0){
                        applyColor(colors, color, i, j, n, m, k);
                        color++;
                    }
                }
            }
            System.out.println(color - 1);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
10 10 3
1 1 1
100 100 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Court Blue (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n=m$ and the time limit is lower. You can make hacks only if both versions of the problem are solved.

In the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.

Let $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:

  * after every round, $\gcd(W_L,W_F)\le 1$;    * at the end of the match, $W_L\le n, W_F\le m$.

Note that $\gcd(0,x)=\gcd(x,0)=x$ for every non-negative integer $x$.

Lelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \cdot W_L + f \cdot W_F$.

Please help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.

Input: The first line contains an integer $t$ ($1\leq t \leq 10^3$) — the number of test cases.

The only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\leq n\leq m \leq 2\cdot 10^7$, $1\leq l,f \leq 10^9$, $\bf{n=m}$): $n$, $m$ gives the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.

Unusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.

Output: For each test case, output a single integer — the maximum total score of a successful performance.

Examples:
input:
8
3 3 2 5
4 4 1 4
6 6 2 2
7 7 2 3
9 9 9 1
2 2 1 4
5 5 1 4
8 8 6 7
output:
19
17
18
33
86
9
24
86

input:
1
20000000 20000000 1341 331
output:
33439999007

input:
2
1984 1984 19 84
9982 9982 44 35
output:
204143
788403

Note: In the first test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Flamm wins, $\gcd(1,2)=1$.    * Flamm wins, $\gcd(1,3)=1$.    * Lelle wins, $\gcd(2,3)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $2\cdot2+3\cdot5=19$.

In the third test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Lelle wins, $\gcd(2,1)=1$.    * Lelle wins, $\gcd(3,1)=1$.    * Lelle wins, $\gcd(4,1)=1$.    * Lelle wins, $\gcd(5,1)=1$.    * Flamm wins, $\gcd(5,2)=1$.    * Flamm wins, $\gcd(5,3)=1$.    * Flamm wins, $\gcd(5,4)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $5\cdot2+4\cdot2=18$. Note that Lelle and Flamm can stop the match even if neither of them has $n$ wins.

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

using namespace std;
#define fastio ios::sync_with_stdio(false), cin.tie(NULL)
#define endl "\n"
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const ll mod = 1e9 + 7;

ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

bool is_prime[2000010];

void solve() {
    ll n, m, l, f;
    cin >> n >> m >> l >> f;
    if(l < f) swap(l, f);

    if(n == 2){
        cout << l * 2 + f << endl;
        return;
    } else if(n == 3){
        cout << l * 3 + f * 2 << endl;
        return;
    } else if(n == 4){
        cout << l * 4 + f << endl;
        return;
    }

    // n이하의 소수 두개 찾기
    ll p1 = 0, p2 = 0;
    for(ll i = n; i >= 2; i--){
        if(is_prime[i]){
            if(p1 == 0) p1 = i;
            else{
                p2 = i;
                break;
            }
        }
    }
    queue<pll> q;
    q.push({p1, p2});
    ll ans = p1 * l + p2 * f;
    while(!q.empty()){
        ll t1 = q.front().first;
        ll t2 = q.front().second;
//        cout << t1 << ' ' << t2 << endl;
        q.pop();
        ans = max(ans, t1 * l + t2 * f);
        if(t1+1 <= n && gcd(t1+1, t2) == 1) {
            q.push({t1+1, t2});
        }
        if(t2+1 <= n && gcd(t1, t2+1) == 1) {
            q.push({t1, t2+1});
        }
    }

    ans = max(ans, n * l + f);
    cout << ans << endl;
}

int main() {
    fastio;
    int TC;
    cin >> TC;
    //TC = 1;

    for(int i = 2; i <= 2000000; i++){
        is_prime[i] = true;
    }

    for(ll i = 2; i <= 2000000; i++){
        if(is_prime[i]){
            for(ll j = i * i; j <= 2000000; j += i){
                is_prime[j] = false;
            }
        }
    }

    while (TC--) {
        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
1
5 5 1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Court Blue (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n=m$ and the time limit is lower. You can make hacks only if both versions of the problem are solved.

In the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.

Let $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:

  * after every round, $\gcd(W_L,W_F)\le 1$;    * at the end of the match, $W_L\le n, W_F\le m$.

Note that $\gcd(0,x)=\gcd(x,0)=x$ for every non-negative integer $x$.

Lelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \cdot W_L + f \cdot W_F$.

Please help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.

Input: The first line contains an integer $t$ ($1\leq t \leq 10^3$) — the number of test cases.

The only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\leq n\leq m \leq 2\cdot 10^7$, $1\leq l,f \leq 10^9$, $\bf{n=m}$): $n$, $m$ gives the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.

Unusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.

Output: For each test case, output a single integer — the maximum total score of a successful performance.

Examples:
input:
8
3 3 2 5
4 4 1 4
6 6 2 2
7 7 2 3
9 9 9 1
2 2 1 4
5 5 1 4
8 8 6 7
output:
19
17
18
33
86
9
24
86

input:
1
20000000 20000000 1341 331
output:
33439999007

input:
2
1984 1984 19 84
9982 9982 44 35
output:
204143
788403

Note: In the first test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Flamm wins, $\gcd(1,2)=1$.    * Flamm wins, $\gcd(1,3)=1$.    * Lelle wins, $\gcd(2,3)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $2\cdot2+3\cdot5=19$.

In the third test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Lelle wins, $\gcd(2,1)=1$.    * Lelle wins, $\gcd(3,1)=1$.    * Lelle wins, $\gcd(4,1)=1$.    * Lelle wins, $\gcd(5,1)=1$.    * Flamm wins, $\gcd(5,2)=1$.    * Flamm wins, $\gcd(5,3)=1$.    * Flamm wins, $\gcd(5,4)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $5\cdot2+4\cdot2=18$. Note that Lelle and Flamm can stop the match even if neither of them has $n$ wins.

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

using namespace std;
using ll = long long;
using pii = pair<int, int>;
using tii = tuple<int, int, int>;
using tll = tuple<ll, ll, ll>;
using ld = long double;
using vi = vector<int>;
using vl = vector<ll>;

#define mp make_pair
#define pb push_back
#define eb emplace_back
#define nl "\n"
#define all(v) v.begin(),v.end()
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define sz(x) (int)(x).size()

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
constexpr int maxn = 2e7 + 5;

ll n, m, q, k, t, a, b, c;
constexpr ll N = 105;

ll hashh(ll a, ll b)
{
	return max(a, b) * 1e9 + min(a, b);
}

unordered_map<ll, bool> can;
bool vis[N][N];
ll omax[N];

ll gcd(ll a, ll b)
{
	if (a > b)
		return gcd(b, a);
	if (a == 0)
		return b;
	return gcd(b % a, a);
}

void dfs(ll a, ll b)
{
	if (a >= N || b >= N)
		return;
	if (vis[a][b])
		return;
	vis[a][b] = true;
	if (gcd(a + 1, b) <= 1)
		dfs(a + 1, b);
	if (gcd(a, b + 1) <= 1)
		dfs(a, b + 1);
}

bool comp[maxn];
vector<ll> primes;
unordered_map<ll, ll> otmax;

void dfs2(ll a, ll b, ll lim)
{
	if (a > b) return;
	if (b > lim) return;
	if (gcd(a, b) > 1)
		return;

	ll h = hashh(a, b);
	if (can[h])
		return;
	can[h] = true;

	otmax[b] = max(otmax[b], a);
	dfs2(a + 1, b, lim);
	dfs2(a, b + 1, lim);
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	dfs(0, 0);

	for (int i = 0; i < N; i++)
	{
		for (int j = 0; j <= i; j++)
		{
			if (vis[i][j])
				omax[i] = j;
		}
	}

	for (ll i = 2; i < maxn; i++)
	{
		if (comp[i]) continue;

		primes.pb(i);
		for (ll j = i * i; j < maxn; j += i)
		{
			comp[j] = true;
		}
	}

	// cout << omax[3];
	// return 0;

	cin >> t;
	while (t--)
	{
		ll n, m, l, f;
		cin >> n >> m >> l >> f;
		if (l > f) swap(l, f);
		if (n < N)
		{
			ll ans = 0;
			for (ll i = 0; i <= n; i++)
			{
				ans = max(ans, l * omax[i] + f * i);
			}
			cout << ans << nl;
			continue;
		}
		auto it = upper_bound(all(primes), n) - primes.begin();
		auto p1 = primes[it - 1];
		auto p2 = primes[it - 2];
		dfs2(p2, p1, n);

		ll ans = 0;
		for (ll i = p1; i <= n; i++)
		{
			ans = max(ans, otmax[i] * l + i * f);
		}
		cout << ans << nl;
	}
}

**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 3 2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Court Blue (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n=m$ and the time limit is lower. You can make hacks only if both versions of the problem are solved.

In the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.

Let $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:

  * after every round, $\gcd(W_L,W_F)\le 1$;    * at the end of the match, $W_L\le n, W_F\le m$.

Note that $\gcd(0,x)=\gcd(x,0)=x$ for every non-negative integer $x$.

Lelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \cdot W_L + f \cdot W_F$.

Please help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.

Input: The first line contains an integer $t$ ($1\leq t \leq 10^3$) — the number of test cases.

The only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\leq n\leq m \leq 2\cdot 10^7$, $1\leq l,f \leq 10^9$, $\bf{n=m}$): $n$, $m$ gives the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.

Unusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.

Output: For each test case, output a single integer — the maximum total score of a successful performance.

Examples:
input:
8
3 3 2 5
4 4 1 4
6 6 2 2
7 7 2 3
9 9 9 1
2 2 1 4
5 5 1 4
8 8 6 7
output:
19
17
18
33
86
9
24
86

input:
1
20000000 20000000 1341 331
output:
33439999007

input:
2
1984 1984 19 84
9982 9982 44 35
output:
204143
788403

Note: In the first test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Flamm wins, $\gcd(1,2)=1$.    * Flamm wins, $\gcd(1,3)=1$.    * Lelle wins, $\gcd(2,3)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $2\cdot2+3\cdot5=19$.

In the third test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Lelle wins, $\gcd(2,1)=1$.    * Lelle wins, $\gcd(3,1)=1$.    * Lelle wins, $\gcd(4,1)=1$.    * Lelle wins, $\gcd(5,1)=1$.    * Flamm wins, $\gcd(5,2)=1$.    * Flamm wins, $\gcd(5,3)=1$.    * Flamm wins, $\gcd(5,4)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $5\cdot2+4\cdot2=18$. Note that Lelle and Flamm can stop the match even if neither of them has $n$ wins.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define db double
#define INF 1000000000
#define inf 100000000000000000ll
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x)
#define putl_(x) printf("%lld ",x)
#define get(x) x=read()
#define putl(x) printf("%lld\n",x)
#define rep(p,n,i) for(int i=p;i<=n;i+=1)
#define fep(n,p,i) for(int i=n;i>=p;--i)
#define go(x) for(int i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define pii pair<int,int>
#define mk make_pair
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define ui unsigned
#define sq sqrt
#define x(w) t[w].x
#define r(w) t[w].r
#define id(w) t[w].id
#define R(w) s[w].r
#define sum(w) t[w].sum
#define sc(A) scanf("%d",&A)
#define scl(A) scanf("%lld",&A)
#define scs(A) scanf("%s",A);
#define put(A) printf("%d\n",A)
#define min(x,y) (x>=y?y:x)
#define max(x,y) (x>=y?x:y)
#define sub(x,y) (x-y<0?x-y+mod:x-y)
#define uint unsigned int
#define mod 998244353
using namespace std;
const int MAXN=20000010,G=3;

int n;
int T;
ll L,F,ans;
int maxx=20000000;
int p[MAXN],v[MAXN],top;
signed main()
{
	sc(T);
	rep(2,maxx,i)
	{
		if(!v[i])
		{
			p[++top]=i;
			v[i]=i;
		}
		rep(1,top,j)
		{
			if(i>maxx/p[j]||v[i]<p[j])break;
			v[i*p[j]]=v[i];
		}
	}
	while(T--)
	{
		sc(n);sc(n);
		scl(L);scl(F);
		if(L<F)swap(L,F);
		ans=0;
		ans=max(ans,F+L*n);
		int l=1,r=top;
		while(l+1<r)
		{
			int mid=(l+r)>>1;
			if(p[mid]<=n)l=mid;
			else r=mid;
		}
		if(p[r]<=n)l=r;
		ans=max(ans,p[l]*L+(p[l]-1)*F);

		if(l>=2)
		{
			ans=max(ans,p[l-1]*F+min(n,p[l-1]*2-1)*L);
		}

		putl(ans);
	}

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to consider:

1. **Edge Cases for Small Values of \( n \)**:
   - Test with the smallest possible values of \( n \) (i.e., \( n = 2 \)).
   - Ensure that \( l \) and \( f \) are such that the gcd condition is critical.

2. **Large Values of \( n \)**:
   - Test with the maximum possible values of \( n \) (i.e., \( n = 2 \times 10^7 \)).
   - Ensure that \( l \) and \( f \) are large to maximize the score and test the performance of the algorithm.

3. **Intermediate Values of \( n \)**:
   - Test with intermediate values to ensure the algorithm handles a range of inputs.
   - Include cases where \( l \) and \( f \) are equal and cases where they are significantly different.

4. **Prime and Composite Values of \( n \)**:
   - Test with prime values of \( n \) to ensure the algorithm handles prime numbers correctly.
   - Test with composite values of \( n \) to ensure the algorithm handles composite numbers correctly.

5. **Special Cases**:
   - Test with \( l = 1 \) and \( f = 1 \) to ensure the algorithm handles the simplest case.
   - Test with \( l \) and \( f \) being large and very different to ensure the algorithm handles extreme cases.

Based on these considerations, here is a small-scale test input that is complete and valid:

```plaintext
5
2 2 1 1
2 2 1 4
3 3 2 5
10 10 1000000000 1
20000000 20000000 1341 331
```

This test input includes:
- Small values of \( n \) (2 and 3).
- Large values of \( n \) (10 and \( 2 \times 10^7 \)).
- Different values of \( l \) and \( f \) to ensure the gcd condition is critical.
- A case with \( l \) and \( f \) being equal.
- A case with \( l \) and \( f \) being very different.

This should help in exposing any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Court Blue (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n=m$ and the time limit is lower. You can make hacks only if both versions of the problem are solved.

In the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.

Let $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:

  * after every round, $\gcd(W_L,W_F)\le 1$;    * at the end of the match, $W_L\le n, W_F\le m$.

Note that $\gcd(0,x)=\gcd(x,0)=x$ for every non-negative integer $x$.

Lelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \cdot W_L + f \cdot W_F$.

Please help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.

Input: The first line contains an integer $t$ ($1\leq t \leq 10^3$) — the number of test cases.

The only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\leq n\leq m \leq 2\cdot 10^7$, $1\leq l,f \leq 10^9$, $\bf{n=m}$): $n$, $m$ gives the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.

Unusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.

Output: For each test case, output a single integer — the maximum total score of a successful performance.

Examples:
input:
8
3 3 2 5
4 4 1 4
6 6 2 2
7 7 2 3
9 9 9 1
2 2 1 4
5 5 1 4
8 8 6 7
output:
19
17
18
33
86
9
24
86

input:
1
20000000 20000000 1341 331
output:
33439999007

input:
2
1984 1984 19 84
9982 9982 44 35
output:
204143
788403

Note: In the first test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Flamm wins, $\gcd(1,2)=1$.    * Flamm wins, $\gcd(1,3)=1$.    * Lelle wins, $\gcd(2,3)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $2\cdot2+3\cdot5=19$.

In the third test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Lelle wins, $\gcd(2,1)=1$.    * Lelle wins, $\gcd(3,1)=1$.    * Lelle wins, $\gcd(4,1)=1$.    * Lelle wins, $\gcd(5,1)=1$.    * Flamm wins, $\gcd(5,2)=1$.    * Flamm wins, $\gcd(5,3)=1$.    * Flamm wins, $\gcd(5,4)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $5\cdot2+4\cdot2=18$. Note that Lelle and Flamm can stop the match even if neither of them has $n$ wins.

Buggy Code:
import math
gcd=math.gcd
x=int(input())
def score(wl,wf,l,f):
    return (l*wl+f*wf)
for i in range(x):
    s=input()
    p=s.split()
    n=int(p[0])
    m=int(p[1])
    l=int(p[2])
    f=int(p[3])
    c=[[0,0],[0,1],[1,0]]
    d=[]
    for i in range(1,n+1):
        for j in range(1,m+1):
            if [i-1,j] not in c and [i,j-1] not in c:
                continue
            if gcd(i,j)==0 or gcd(i,j)==1:
                z=[i,j]
                c.append(z)
    for i in c:
        d.append(score(i[0],i[-1],l,f))
    print(max(d))

**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 1000000000 1
2 2 1 1000000000
3 3 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Court Blue (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n=m$ and the time limit is lower. You can make hacks only if both versions of the problem are solved.

In the court of the Blue King, Lelle and Flamm are having a performance match. The match consists of several rounds. In each round, either Lelle or Flamm wins.

Let $W_L$ and $W_F$ denote the number of wins of Lelle and Flamm, respectively. The Blue King considers a match to be successful if and only if:

  * after every round, $\gcd(W_L,W_F)\le 1$;    * at the end of the match, $W_L\le n, W_F\le m$.

Note that $\gcd(0,x)=\gcd(x,0)=x$ for every non-negative integer $x$.

Lelle and Flamm can decide to stop the match whenever they want, and the final score of the performance is $l \cdot W_L + f \cdot W_F$.

Please help Lelle and Flamm coordinate their wins and losses such that the performance is successful, and the total score of the performance is maximized.

Input: The first line contains an integer $t$ ($1\leq t \leq 10^3$) — the number of test cases.

The only line of each test case contains four integers $n$, $m$, $l$, $f$ ($2\leq n\leq m \leq 2\cdot 10^7$, $1\leq l,f \leq 10^9$, $\bf{n=m}$): $n$, $m$ gives the upper bound on the number of Lelle and Flamm's wins, $l$ and $f$ determine the final score of the performance.

Unusual additional constraint: it is guaranteed that, for each test, there are no pairs of test cases with the same pair of $n$, $m$.

Output: For each test case, output a single integer — the maximum total score of a successful performance.

Examples:
input:
8
3 3 2 5
4 4 1 4
6 6 2 2
7 7 2 3
9 9 9 1
2 2 1 4
5 5 1 4
8 8 6 7
output:
19
17
18
33
86
9
24
86

input:
1
20000000 20000000 1341 331
output:
33439999007

input:
2
1984 1984 19 84
9982 9982 44 35
output:
204143
788403

Note: In the first test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Flamm wins, $\gcd(1,2)=1$.    * Flamm wins, $\gcd(1,3)=1$.    * Lelle wins, $\gcd(2,3)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $2\cdot2+3\cdot5=19$.

In the third test case, a possible performance is as follows:

  * Flamm wins, $\gcd(0,1)=1$.    * Lelle wins, $\gcd(1,1)=1$.    * Lelle wins, $\gcd(2,1)=1$.    * Lelle wins, $\gcd(3,1)=1$.    * Lelle wins, $\gcd(4,1)=1$.    * Lelle wins, $\gcd(5,1)=1$.    * Flamm wins, $\gcd(5,2)=1$.    * Flamm wins, $\gcd(5,3)=1$.    * Flamm wins, $\gcd(5,4)=1$.    * Lelle and Flamm agree to stop the match.

The final score is $5\cdot2+4\cdot2=18$. Note that Lelle and Flamm can stop the match even if neither of them has $n$ wins.

Buggy Code:
import sys
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
def rec(x, y):
    global ans
    ans = max(ans, x * l + y * f)
    if math.gcd(x + 1, y) == 1 and x + 1 <= n:
        rec(x + 1, y)
    if math.gcd(x, y + 1) == 1 and y + 1 <= n:
        rec(x, y + 1)
G = 10 ** 7 + 3
p = [1] * G
for i in range(2, G):
    for j in range(2 * i, G, i):
        p[j] = 0
F = [2] * G
for i in range(3, G):
    if p[i]:
        F[i] = i
    else:
        F[i] = F[i - 1]
for __ in range(fg()):
    n, m, l, f = fgh()
    if l > f:
        l, f = f, l
    p = F[n]
    ans = (p - 1) * l + p * f
    for j in range(p - 1, F[p - 1] - 2, -1):
        rec(j, p)
    print(ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code seems to have issues with handling large numbers and the recursive function `rec` might not terminate correctly in certain cases. Let's consider the following aspects:

1. **Small and Simple Cases**: Test the code with small values to ensure it handles basic scenarios correctly.
2. **Large Values**: Test with large values to check if the code can handle them efficiently.
3. **Equal Values for l and f**: Test with equal values for `l` and `f` to see how the code handles symmetry.
4. **Prime Values**: Test with `n` and `m` being prime numbers to see if the code correctly handles the constraints.
5. **High Contrast Values for l and f**: Test with very different values for `l` and `f` to see how the code balances the wins.

Given these considerations, here is a small-scale test input that should expose potential bugs:

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

This input includes:
- A simple case where `n = m = 3` and `l = 2`, `f = 5`.
- A case where `n = m = 4` and `l = 1`, `f = 4`.
- A symmetric case where `n = m = 6` and `l = 2`, `f = 2`.
- A case where `n = m = 7` and `l = 2`, `f = 3`.
- A case with high contrast values where `n = m = 9` and `l = 9`, `f = 1`.

These cases should help in identifying any issues with the recursive function and the handling of large numbers.