**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
import com.sun.source.tree.BinaryTree;

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

public class Ez {

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

  public static void main(String[] args) throws IOException {
    new Thread(null, new Solution(), "solution", 1 << 26).start();
  }

  private static class Solution implements Runnable {

    boolean[] prime;

    public void run() {
      int testCases = sc.nextInt();
      //int testCases = 1;
      for (int i = 1; i <= testCases; ++i) {
        solve();
      }
      out.flush();
    }

    private void solve() {
      int n = sc.nextInt();
      long k = sc.nextLong();
      int[] a = sc.readArray(n);
      long sum = 0;
      long min = Integer.MAX_VALUE;
      Arrays.sort(a);
      long[] pref = new long[n + 1];
      for (int i = 1; i <= n; i++) {
        pref[i] = pref[i - 1] + a[i - 1];
      }
      for (int i = 0; i <= n; i++) {
        for (int j = n ; j > i; j--) {
          if (pref[j] - pref[i] <= k) {
            min = Math.min(min, k - (pref[j] - pref[i]));
          }
        }
      }
      out.println(min);
    }

    long gcd(long a, long b) {
      return b == 0 ? a : gcd(b, a % b);
    }

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

    private void yN(boolean flag) {
      out.println(flag ? "YES" : "NO");
    }

    private void getPrimes() {
      long N = 100000;
      prime = new boolean[(int) N + 1];
      prime[0] = true;
      prime[1] = true;
      for (long i = 2; i <= N; i++) {
        if (!prime[(int) i]) {
          if (i * i <= N) {
            for (long j = i * i; j <= N; j += i) {
              prime[(int) j] = true;
            }
          }
        }
      }
    }

    boolean checkAns(int n, long mid, int k, Map<Integer, Integer> out, Map<Integer, Integer> in) {
      long max = (long) k * mid;
      for (int i = 0; i < n; i++) {
        max += in.get(i);
        max -= out.get(i);
        if (max < 0) {
          return false;
        }
      }
      return true;
    }

    int lower(int[] a, int n, int target) {
      int l = -1, r = n, mid;
      while (l + 1 < r) {
        mid = (l + r) / 2;
        if (a[mid] < target) {
          l = mid;
        } else {
          r = mid;
        }
      }
      return l;
    }

    boolean check(long[] a, int n, long k) {
      boolean flag = false;
      long last = a[0];
      for (int i = 1; i < n; i++) {
        if (a[i] - last <= k) {
          last = a[i];
          continue;
        }
        if (!flag) {
          last = a[i];
          flag = true;
        } else {
          return false;
        }
      }
      return true;
    }

    int find(int[] a, int n) {
      int cnt = 0;
      for (int i = 1; i < n; i++) {
        if (a[i - 1] == a[i]) {
          cnt++;
        }
      }
      return cnt;
    }

    int minTravel(int r, int c, int[][] a, int n) {
      int min = Integer.MAX_VALUE;
      while (r < n && c < n) {
        min = Math.min(min, a[r][c]);
        r++;
        c++;
      }
      return min;
    }

    boolean ask(String s) {
      out.println("? " + s);
      out.flush();
      return sc.nextInt() == 1;
    }

    private boolean isTrue(long sum, int n, long[] a) {
      double midVal = (double) sum / n / 2;
      int l = -1, r = n, mid;
      while (l + 1 < r) {
        mid = (l + r) / 2;
        if (midVal > a[mid]) {
          l = mid;
        } else {
          r = mid;
        }
      }
      return l >= (n / 2);
    }

    private long getVal(long num) {
      long min = 0, max = Integer.MAX_VALUE, mid;
      while (max - min > 1) {
        mid = (max + min) >> 1;
        if (mid * mid <= num) {
          min = mid;
        } else {
          max = mid;
        }
      }
      return min;
    }
  }

  private static class IOHandler {
    BufferedReader br;
    StringTokenizer st;

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

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

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

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

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

      return res;
    }

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

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

      return res;
    }

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

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

      int a, b;

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

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

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

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

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

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

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

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

      return graph;
    }

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

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

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

  }

}

class Pair {
  int a;
  int b;

  public Pair(int a, int b) {
    this.a = a;
    this.b = b;
  }

  long sum() {
    return a + b;
  }
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to craft a test case that highlights the issues in the logic. The primary issue in the code is the way it calculates the minimum number of coins to add. The code incorrectly uses a double loop to check all possible subarrays, which is inefficient and can lead to incorrect results.

### Step-by-Step Reasoning:

1. **Identify the Bug**: The code uses a nested loop to check all subarrays and calculates the minimum number of coins to add. This approach is not optimal and can lead to incorrect results, especially when the initial sum of coins is already close to `k`.

2. **Edge Cases**:
   - **Initial Sum is Exactly `k`**: The code should return 0 if the initial sum of coins is exactly `k`.
   - **Initial Sum is Just Below `k`**: The code should return the smallest number of coins needed to make the sum exactly `k`.
   - **All Chests Have the Same Number of Coins**: The code should handle cases where all chests have the same number of coins.
   - **Large `k` and Small `n`**: The code should handle cases where `k` is much larger than the sum of the initial coins in the chests.

3. **Construct the Test Case**:
   - **Test Case 1**: Initial sum is exactly `k`.
   - **Test Case 2**: Initial sum is just below `k`.
   - **Test Case 3**: All chests have the same number of coins.
   - **Test Case 4**: Large `k` and small `n`.

### Test Input:

```plaintext
4
3 10
3 3 4
3 11
3 3 4
3 12
4 4 4
3 20
1 1 1
```

### Explanation:

1. **Test Case 1**:
   - `n = 3`, `k = 10`
   - Initial coins: `3, 3, 4`
   - Sum: `3 + 3 + 4 = 10`
   - Expected output: `0` (no coins need to be added)

2. **Test Case 2**:
   - `n = 3`, `k = 11`
   - Initial coins: `3, 3, 4`
   - Sum: `3 + 3 + 4 = 10`
   - Expected output: `1` (1 coin needs to be added to make the sum 11)

3. **Test Case 3**:
   - `n = 3`, `k = 12`
   - Initial coins: `4, 4, 4`
   - Sum: `4 + 4 + 4 = 12`
   - Expected output: `0` (no coins need to be added)

4. **Test Case 4**:
   - `n = 3`, `k = 20`
   - Initial coins: `1, 1, 1`
   - Sum: `1 + 1 + 1 = 3`
   - Expected output: `17` (17 coins need to be added to make the sum 20)

This test input should effectively expose the bug in the provided code by covering a range of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
import java.util.*;

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

        while (t-- > 0) {
            int n = sc.nextInt(); // Number of chests
            int k = sc.nextInt(); // Target coins
            int[] coins = new int[n];

            for (int i = 0; i < n; i++) {
                coins[i] = sc.nextInt();
            }

            // Sort the chests in descending order
            Arrays.sort(coins);
            int[] sortedCoins = new int[n];
            for (int i = 0; i < n; i++) {
                sortedCoins[i] = coins[n - i - 1];
            }

            int currentSum = 0;
            int addedCoins = 0;

            for (int i = 0; i <n; i++) {
                if (currentSum + sortedCoins[i] >= k) {
                    // Add enough coins to make the current sum exactly k
                    if(i==0) {
                        addedCoins+=Math.max(0,k-currentSum-sortedCoins[i]);
                        break;
                    }
                    addedCoins += Math.max(0, k - currentSum);
                    break;
                } else {
                    currentSum += sortedCoins[i];
                    if(i+1==n) {
                        addedCoins += k - currentSum;
                    }
                }
            }

            System.out.println(addedCoins);
        }

        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
1
5 15
5 5 5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

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

using namespace std;

#define int long long
#define f first
#define s second
#define int long long
#define vi  vector<int>
#define input(arr) for(int i = 0; i < arr.size(); i++) cin >> arr[i];
#define output(a) for(int i = 0; i < a.size(); i++) {cout << a[i] << " ";}
#define yes cout << "yes" << '\n';
#define no cout << "no" << '\n';
#define nl cout << '\n';
#define pb push_back
#define op(a) cout << a << '\n';

int32_t main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int T;

    cin >> T;

    while(T--) {

      int n, k;
      cin >> n >> k;

      vi arr(n);
      input(arr);

      sort(arr.begin(), arr.end(), greater<int>());

      int s = 0;

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

        s += arr[i];
      }
      int l = 0;

      int sum = 0;
      for(int i = 0; i < n; i++) {

        sum += arr[i];

        if(sum > k) {

          cout << k - (sum - arr[i - 1] ) << '\n';
          break;
        }

        if(sum == k) {

          cout << 0 << '\n';
          break;

        }

        if(i == n - 1) {

          l = 1;
          break;
        }

      }

      if( l == 1) {

        cout << k - s << '\n';

      }

    }

    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 15
5 5 5 5 5
```

**Explanation:**
- This test case is designed to expose the bug in the provided code by setting all chest values to the same number, which is a common edge case.
- The total number of coins in the chests is 25, which is more than the required 15 coins.
- The code should ideally output 0 because Monocarp can take the first three chests (5 + 5 + 5 = 15) to get exactly 15 coins without needing to add any additional coins.
- However, the buggy code might incorrectly calculate the number of coins to add due to its logic of subtracting the sum of the first few chests from `k` when the sum exceeds `k`. This can lead to incorrect results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

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

#define fast_io ios::sync_with_stdio(false); cin.tie(0);
#define pb push_back
#define all(v) v.begin(), v.end()

using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;

const int MOD = 1e9 + 7;

void solve() {
	ll n, k;
	cin >> n >> k;
	vector<ll> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	sort(a.rbegin(), a.rend());
	ll sum = 0;
	ll i;
	for (i = 0; sum + a[i] <= k; i++) {
		sum += a[i];
	}
	cout << k - sum << "\n";
}

int main() {
	fast_io;

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

	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on scenarios where the code might fail to handle the problem correctly. The main issue in the provided code is that it does not account for the minimum number of coins needed to be added to make the total sum of coins in the chests exactly `k` when Monocarp stops taking the chests.

### Steps to Identify the Bug:
1. **Initial Sum Calculation**: The code correctly calculates the initial sum of the coins in the chests.
2. **Greedy Selection**: The code correctly sorts the chests in descending order and iterates through them to find the point where the sum of the coins in the selected chests is just less than or equal to `k`.
3. **Output Calculation**: The code outputs `k - sum`, which is the difference between `k` and the sum of the coins in the selected chests. However, this approach does not ensure that the minimum number of coins is added to make the total sum exactly `k`.

### Challenging Test Input:
To expose the bug, we can create a test case where the initial sum of the coins in the chests is just less than `k`, and adding the minimum number of coins to any chest would make the total sum exactly `k`.

### Example Test Input:
```plaintext
1
5 15
4 4 3 2 1
```

### Explanation:
- **n = 5**: There are 5 chests.
- **k = 15**: The total number of coins in the chests must be at least 15.
- **a = [4, 4, 3, 2, 1]**: The initial number of coins in each chest.

### Expected Output:
- The minimum number of coins to be added to make the total sum exactly 15 is 2. This can be achieved by adding 1 coin to the first chest and 1 coin to the second chest.

### Why This Test Input is Challenging:
- The initial sum of the coins in the chests is 14.
- The code will correctly sort the chests and sum the coins in the first few chests.
- However, the code will output `15 - 14 = 1`, which is incorrect because adding just 1 coin is not enough to make the total sum exactly 15. The correct answer is 2.

This test input should expose the bug in the provided code and highlight the need to ensure that the minimum number of coins is added to make the total sum exactly `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
import sys
from os import path
from math import gcd,floor,sqrt,log,ceil,factorial
from bisect import bisect_left, bisect_right
import heapq
from collections import defaultdict, deque
from  itertools import combinations
############ ---- Input Functions ---- ####################
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def inst():
    return(list(map(str,input().split())))
def insr():
    s = input().strip()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
def inmatrix(n,m):
    arr=[]
    for i in range(n):
        temp=inlt()
        arr.append(temp)
    return arr
############################################################
def cuberoot(x):
    if 0<=x: return x**(1./2.)
    return -(-x)**(1./2.)
def is_anagram(str1, str2):
    return sorted(str1) == sorted(str2)
def odd_composite_numbers_optimized(n):
    # Sieve of Eratosthenes
    is_prime = [True] * (n + 1)
    is_prime[0] = is_prime[1] = False
    for i in range(2, int(n**0.5) + 1):
        if is_prime[i]:
            for j in range(i * i, n + 1, i):
                is_prime[j] = False

    # Find odd composite numbers
    return [i for i in range(9, n + 1, 2) if not is_prime[i]]
############################################################
MOD=1000000007
INF=float("Inf")
MAX=1000000
############################################################
if(path.exists('input.txt')):
    sys.stdin = open("input.txt","r")
    sys.stdout = open("output.txt","w")
    sys.stderr = open("error.txt", "w")
else:
    input = sys.stdin.readline
############################################################
def find_or_closest(arr, i):
    """
    Find the element `i` in a sorted array `arr`.
    If not present, return the smallest closest value.
    """
    left, right = 0, len(arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == i:
            return i
        elif arr[mid] < i:
            left = mid + 1
        else:
            right = mid - 1

    # At this point, `left` is the index of the smallest element larger than `i`
    # and `right` is the index of the largest element smaller than `i`.
    if left >= len(arr):  # `i` is greater than all elements
        return arr[right]
    elif right < 0:  # `i` is smaller than all elements
        return arr[left]
    else:
        # Choose the closest value
        return arr[left] if abs(arr[left] - i) < abs(arr[right] - i) else arr[right]
##################################################################

def solve():
    n,k=invr()
    a=inlt()
    a.sort(reverse=True)
    ss=sum(a)
    if ss<k:
        print(k-ss)
        return
    ass=0
    for i in range(n):
        ass+=a[i]
        if ass==k:
            print(0)
            return
        if ass>k:
            ass-=a[i-1]
            print(k-ass)
            return

def main():
    t=1
    t=inp()
    # print(t)
    while(t):
        t-=1
        solve()
    #solve()

main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

Buggy Code:
t=int(input())
while t:
    t-=1
    n,k=map(int,input().split())
    l=list(map(int,input().split()))
    l.sort(reverse=True)
    new_lis=[0 for i in range(n)]
    new_lis[0]=l[0]
    for i in range(1,n):
        new_lis[i]=l[i]+new_lis[i-1]

    find_ind=-1
    for i in range(n):
        if new_lis[i]==k:
            find_ind=i
            break
        elif new_lis[i]>k:
            find_ind=i
            break

    if(find_ind==0):
        print(0)
    elif(find_ind>0):
        print(k-new_lis[find_ind-1])
    else:
        print(k-new_lis[n-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
5 10
4 3 2 1 1
3 15
5 5 5
2 100
1 99
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

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

public class EqualizeArray {

    public static boolean isPrime(long n) {
        // Check if n is less than 2
        if (n < 2) {
            return false;
        }

        // Check if n is 2 or 3
        if (n == 2 || n == 3) {
            return true;
        }

        // Eliminate even numbers and multiples of 3
        if (n % 2 == 0 || n % 3 == 0) {
            return false;
        }

        // Check for factors up to sqrt(n)
        for (long i = 5; i * i <= n; i += 6) {
            if (n % i == 0 || n % (i + 2) == 0) {
                return false;
            }
        }

        return true;
    }

    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);
        int N = s.nextInt();

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

            int n = s.nextInt();
            int hark = s.nextInt();
            int[] arr = new int[n];

            for (int i = 0; i < n; i++) {
                arr[i] = s.nextInt();
            }
            Arrays.sort(arr);

            int target = hark;
            int ans = 0;

            for( int ij = n-1;ij>=0;ij--){
                int val = arr[ij];
                if( target == val ){
                    target = 0;
                    break;
                }
                if( target > val ){
                    target -= val;
                }
                if( target < val ){
                    ans = target;
                    target = 0;
                    break;
                }
            }
            ans += target;

            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
3
5 15
5 5 5 5 5
3 10
1 2 3
4 20
10 5 4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Greedy Monocarp
time_limit_ms: 2000
memory_limit_mb: 512
Description: There are $$$n$$$ chests; the $$$i$$$-th chest initially contains $$$a_i$$$ coins. For each chest, you can choose any non-negative ($$$0$$$ or greater) number of coins to add to that chest, with one constraint: the total number of coins in all chests must become at least $$$k$$$.

After you've finished adding coins to the chests, greedy Monocarp comes, who wants the coins. He will take the chests one by one, and since he is greedy, he will always choose the chest with the maximum number of coins. Monocarp will stop as soon as the total number of coins in chests he takes is at least $$$k$$$.

You want Monocarp to take as few coins as possible, so you have to add coins to the chests in such a way that, when Monocarp stops taking chests, he will have exactly $$$k$$$ coins. Calculate the minimum number of coins you have to add.

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

Each test case consists of two lines:

- the first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$; $$$1 \le k \le 10^7$$$);
- the second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le k$$$).

Output: For each test case, print one integer — the minimum number of coins you have to add so that, when Monocarp stops taking the chests, he has exactly $$$k$$$ coins. It can be shown that under the constraints of the problem, it is always possible.

Examples:
input:
4
5 4
4 1 2 3 2
5 10
4 1 2 3 2
2 10
1 1
3 8
3 3 3
output:
0
1
8
2

Note: In the first test case of the example, you don't have to add any coins. When Monocarp arrives, he will take the chest with $$$4$$$ coins, so he will have exactly $$$4$$$ coins.

In the second test case of the example, you can add $$$1$$$ coin to the $$$4$$$-th chest, so, when Monocarp arrives, he will take a chest with $$$4$$$ coins, then another chest with $$$4$$$ coins, and a chest with $$$2$$$ coins.

In the third test case of the example, you can add $$$3$$$ coins to the $$$1$$$-st chest and $$$5$$$ coins to the $$$2$$$-nd chest.

In the fourth test case of the example, you can add $$$1$$$ coin to the $$$1$$$-st chest and $$$1$$$ coin to the $$$3$$$-rd chest.

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

@SuppressWarnings("unused")

public class Codeforces {

    public static void solve(Scanner scanner) {

        Integer size = scanner.nextInt();
        Integer limit = scanner.nextInt();

        ArrayList<Integer> list = new ArrayList<>();

        for (int i = 0; i < size; i++)
        {
            list.add(scanner.nextInt());
        }

        Collections.sort(list, Collections.reverseOrder());

        if (list.get(0) >= limit)
        {
            System.out.println(0);
            return;
        }

        Integer sum = 0;
        Integer index = -1;

        for (int i = 0; i < list.size(); i++)
        {
            sum += list.get(i);

            if (sum >= limit)
            {
                index = i;
                break;
            }
        }

        Integer end = (index == -1) ? list.size() : index;
        sum = 0;

        for (int i = 0; i < end; i++)
        {
            sum += list.get(i);
        }

        System.out.println(limit - sum);

    }

    public static void main(String args[]) {

        Scanner scanner = new Scanner(System.in);

        int test = scanner.nextInt();
        scanner.nextLine();

        for (int t = 0; t < test; t++)
        {
            solve(scanner);
        }

        scanner.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Carousel of Combinations
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. The function $$$C(i,k)$$$ represents the number of distinct ways you can select $$$k$$$ distinct numbers from the set {$$$1, 2, \ldots, i$$$} and arrange them in a circle$$$^\dagger$$$.

Find the value of $$$$$$ \sum\limits_{i=1}^n \sum\limits_{j=1}^i \left( C(i,j) \bmod j \right). $$$$$$ Here, the operation $$$x \bmod y$$$ denotes the remainder from dividing $$$x$$$ by $$$y$$$.

Since this value can be very large, find it modulo $$$10^9+7$$$.

$$$^\dagger$$$ In a circular arrangement, sequences are considered identical if one can be rotated to match the other. For instance, $$$[1, 2, 3]$$$ and $$$[2, 3, 1]$$$ are equivalent in a circle.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$).

Output: For each test case, output a single integer on a new line — the value of the expression to be calculated modulo $$$10^9+7$$$.

Examples:
input:
4
1
3
6
314159
output:
0
4
24
78926217

Note: In the first test case, $$$C(1,1) \bmod 1 = 0$$$.

In the second test case:

- $$$C(1,1)=1$$$ (the arrangements are: $$$[1]$$$);
- $$$C(2,1)=2$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$);
- $$$C(2,2)=1$$$ (the arrangements are: $$$[1, 2]$$$);
- $$$C(3,1)=3$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$, $$$[3]$$$);
- $$$C(3,2)=3$$$ (the arrangements are: $$$[1, 2]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$);
- $$$C(3,3)=2$$$ (the arrangements are: $$$[1, 2, 3]$$$, $$$[1, 3, 2]$$$).

Buggy Code:
#include<cstdio>
#define ll long long
#define maxn 1000005
using namespace std;

const int M=1000000;
const ll mod=1e9+7,inv2=500000004;
int t,n;
int tot,prime[maxn],mark[maxn];

void clear()
{

}

void ready()
{
	for(int i=2;i<=M;++i)
	{
		if(!mark[i])
		prime[++tot]=i;
		for(int j=1;j<=tot;++j)
		{
			int x=i*prime[j];
			if(x>M)
			break;
			mark[x]=1;
			if(i%prime[j]==0)
			break;
		}
	}
}

ll solve4()
{
	ll ret=0;
	for(int i=0;i<n-3;++i)
	{
		int res=i/4;
		if(res%2==0)
		ret+=2;
	}
	return ret;
}

ll getans(ll x,ll n)
{
	ll cnt1=n/x;
	if(cnt1==0)
	return n*(x-1)%mod;
	if(n%x)
	cnt1++;
	ll cnt2=cnt1/x,res1=n%x,res2=cnt1%x;
	if(res1==0)
	res1=x;
	ll ret=x*x%mod*(x-1)%mod*inv2%mod*cnt2%mod;
	if(res2)
	{
		ret=(ret+x*(res2-1)%mod*(2*x-res2)%mod*inv2%mod)%mod;
		ret=(ret+res1*(x-res2)%mod)%mod;
	}
	return ret;
}

void solve()
{
	ll ans=0;
	scanf("%d",&n);
	ll frac=1;
	for(int i=1;i<=tot;++i)
	{
		ll x=prime[i];
		if(x>n)
		break;
		ll ret=getans(x,n-x+1);
		ans=(ans+ret)%mod;
	}
	if(n>=4)
	ans=(ans+solve4())%mod;
	printf("%lld\n",ans);
}

int main()
{
	ready();
	scanf("%d",&t);
	while(t--)
	{
		solve();
		clear();
	}
	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 where the code might fail. Here are some considerations:

1. **Edge Cases:**
   - Small values of `n` where the behavior of the function can be manually verified.
   - Values of `n` that are just above the threshold of the prime sieve (i.e., around 1,000,000).

2. **Large Values:**
   - Values of `n` that are large enough to test the performance and correctness of the algorithm, but not so large that they exceed the input limits.

3. **Special Values:**
   - Values of `n` that are powers of small primes (e.g., 2, 3, 5) to test the modular arithmetic and prime handling.

Given these considerations, let's create a test input that includes a mix of these cases:

```plaintext
6
1
3
4
1000000
1000001
314159
```

This input includes:
- Small values (1, 3, 4) to verify basic behavior.
- A value just above the prime sieve limit (1,000,000 and 1,000,001) to test the edge of the prime sieve.
- A large value (314,159) to test the performance and correctness for larger inputs.

This should help in exposing any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Carousel of Combinations
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. The function $$$C(i,k)$$$ represents the number of distinct ways you can select $$$k$$$ distinct numbers from the set {$$$1, 2, \ldots, i$$$} and arrange them in a circle$$$^\dagger$$$.

Find the value of $$$$$$ \sum\limits_{i=1}^n \sum\limits_{j=1}^i \left( C(i,j) \bmod j \right). $$$$$$ Here, the operation $$$x \bmod y$$$ denotes the remainder from dividing $$$x$$$ by $$$y$$$.

Since this value can be very large, find it modulo $$$10^9+7$$$.

$$$^\dagger$$$ In a circular arrangement, sequences are considered identical if one can be rotated to match the other. For instance, $$$[1, 2, 3]$$$ and $$$[2, 3, 1]$$$ are equivalent in a circle.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$).

Output: For each test case, output a single integer on a new line — the value of the expression to be calculated modulo $$$10^9+7$$$.

Examples:
input:
4
1
3
6
314159
output:
0
4
24
78926217

Note: In the first test case, $$$C(1,1) \bmod 1 = 0$$$.

In the second test case:

- $$$C(1,1)=1$$$ (the arrangements are: $$$[1]$$$);
- $$$C(2,1)=2$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$);
- $$$C(2,2)=1$$$ (the arrangements are: $$$[1, 2]$$$);
- $$$C(3,1)=3$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$, $$$[3]$$$);
- $$$C(3,2)=3$$$ (the arrangements are: $$$[1, 2]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$);
- $$$C(3,3)=2$$$ (the arrangements are: $$$[1, 2, 3]$$$, $$$[1, 3, 2]$$$).

Buggy Code:
#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

typedef long long Long;

const Long mod = 1000000007;
const Long nmax = 1000000;

bool sv[nmax+1];
vector<Long> pp;
vector<Long> pu;

void init() {
    for(Long p=0; p<=nmax; p++) {
        sv[p] = true;
    }
    sv[0] = sv[1] = false;
    for(Long p=0; p<=nmax; p++) {
        if(sv[p]) {
            pp.push_back(p);
            for(Long k=p*p; k<=nmax; k+=p) {
                sv[k] = false;
            }
        }
    }
}

Long solve(Long n, bool foo) {
    auto d0 = div(n+1, 4ll);
    auto d1 = div(d0.quot, 2ll);
    Long s = 8*d1.quot;
    if(d1.rem) {
        s += 2*d0.rem;
    }
    for(int i=0; i<pp.size(); i++) {
        Long p = pp[i];
        if(p>n) {
            break;
        }
        if(foo) {
            Long k = (n+1)/p;
            Long r = (n+1)%p;
            Long u = k/p;
            k %= p;
            Long v = pu[i]*u;
            s += v;
            continue;
        }
        d0 = div(n+1, p);
        d1 = div(d0.quot, p);
        Long v = pu[i]*d1.quot; // p*(p-1)/2*p*d1.quot;
        if(d1.rem) {
            v += (d1.rem-1)*(2*p-d1.rem)/2 %mod *p;
            v += d0.rem*(p-d1.rem);
        };
        s += v;
        // s %= mod;
    }
    return s % mod;
}

int main() {
    init();
    for(Long p: pp) {
        pu.push_back(p*(p-1)/2 %mod *p %mod);
    }
    int t;
    cin >> t;
    for(int i=0; i<t; i++) {
        Long n;
        cin >> n;
        cout << solve(n, t==100000) << '\n';
    }
    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
1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Carousel of Combinations
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. The function $$$C(i,k)$$$ represents the number of distinct ways you can select $$$k$$$ distinct numbers from the set {$$$1, 2, \ldots, i$$$} and arrange them in a circle$$$^\dagger$$$.

Find the value of $$$$$$ \sum\limits_{i=1}^n \sum\limits_{j=1}^i \left( C(i,j) \bmod j \right). $$$$$$ Here, the operation $$$x \bmod y$$$ denotes the remainder from dividing $$$x$$$ by $$$y$$$.

Since this value can be very large, find it modulo $$$10^9+7$$$.

$$$^\dagger$$$ In a circular arrangement, sequences are considered identical if one can be rotated to match the other. For instance, $$$[1, 2, 3]$$$ and $$$[2, 3, 1]$$$ are equivalent in a circle.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$).

Output: For each test case, output a single integer on a new line — the value of the expression to be calculated modulo $$$10^9+7$$$.

Examples:
input:
4
1
3
6
314159
output:
0
4
24
78926217

Note: In the first test case, $$$C(1,1) \bmod 1 = 0$$$.

In the second test case:

- $$$C(1,1)=1$$$ (the arrangements are: $$$[1]$$$);
- $$$C(2,1)=2$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$);
- $$$C(2,2)=1$$$ (the arrangements are: $$$[1, 2]$$$);
- $$$C(3,1)=3$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$, $$$[3]$$$);
- $$$C(3,2)=3$$$ (the arrangements are: $$$[1, 2]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$);
- $$$C(3,3)=2$$$ (the arrangements are: $$$[1, 2, 3]$$$, $$$[1, 3, 2]$$$).

Buggy Code:
#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
using ll = long long;
constexpr int N = 1e6 + 10;
constexpr ll P = 1e9 + 7;
bool st[N];
int primes[N],cnt;
void init(int n){
    for(int i = 2;i < n;i ++){
        if(!st[i]) primes[cnt ++] = i;
        for(int j = 0;primes[j] < n/i;j ++){
            st[i*primes[j]] = true;
            if(i % primes[j] == 0) break;
        }
    }
}
void solve(){
    ll n;
    cin >> n;
    ll ans = 0;
    for(int j = 0;primes[j] <= n;j ++){
        ll t = primes[j];
        ll cnt = 1;
        for(int l = t;l <= n;l += t){
            if(cnt == 0){
                cnt ++;
                continue;
            }
            ll r = min(n,l + t - 1);
            ans = (ans + (r-l+1) * (t-cnt) ) % P;
            // cout << l << " " << r << " " << t - cnt << "\n";
            cnt = (cnt + 1) % t;

        }
        // cout << t << " " << ans << "\n";
    }
    if(n >= 4){
        ll cnt = 1;
        ll t = 4;
        for(int l = 4;l <= n;l += t){
            ll r = min(n,l + t - 1);
            ans = (ans + (r-l+1) * (cnt*2%4)) % P;
            cnt ++;
        }
    }
    cout << ans << "\n";

}
int main(){
    init(N);
    int t;
    cin >> t;
    while(t --) solve();
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
1
4
10
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Carousel of Combinations
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. The function $$$C(i,k)$$$ represents the number of distinct ways you can select $$$k$$$ distinct numbers from the set {$$$1, 2, \ldots, i$$$} and arrange them in a circle$$$^\dagger$$$.

Find the value of $$$$$$ \sum\limits_{i=1}^n \sum\limits_{j=1}^i \left( C(i,j) \bmod j \right). $$$$$$ Here, the operation $$$x \bmod y$$$ denotes the remainder from dividing $$$x$$$ by $$$y$$$.

Since this value can be very large, find it modulo $$$10^9+7$$$.

$$$^\dagger$$$ In a circular arrangement, sequences are considered identical if one can be rotated to match the other. For instance, $$$[1, 2, 3]$$$ and $$$[2, 3, 1]$$$ are equivalent in a circle.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$).

Output: For each test case, output a single integer on a new line — the value of the expression to be calculated modulo $$$10^9+7$$$.

Examples:
input:
4
1
3
6
314159
output:
0
4
24
78926217

Note: In the first test case, $$$C(1,1) \bmod 1 = 0$$$.

In the second test case:

- $$$C(1,1)=1$$$ (the arrangements are: $$$[1]$$$);
- $$$C(2,1)=2$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$);
- $$$C(2,2)=1$$$ (the arrangements are: $$$[1, 2]$$$);
- $$$C(3,1)=3$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$, $$$[3]$$$);
- $$$C(3,2)=3$$$ (the arrangements are: $$$[1, 2]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$);
- $$$C(3,3)=2$$$ (the arrangements are: $$$[1, 2, 3]$$$, $$$[1, 3, 2]$$$).

Buggy Code:
import os, sys
from io import BytesIO, IOBase

from collections import defaultdict, deque, Counter
from bisect import bisect_left, bisect_right
from heapq import heappush, heappop
from functools import lru_cache, reduce
from itertools import accumulate
from operator import xor
import random
import math
import sys

# sys.setrecursionlimit(10 ** 6)
# Fast IO Region
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
sys.setrecursionlimit(4000)

ii = lambda: int(input())
mii = lambda: map(int, input().split())
fii = lambda: map(float, input().split())
lmii = lambda: list(map(int, input().split()))
i2c = lambda n: chr(ord('a') + n)
c2i = lambda c: ord(c) - ord('a')

# mod = 998244353

mod = 10 ** 9 + 7

flush = lambda: sys.stdout.flush()

P = [1] * (10 ** 6 + 1)
primes = []
for i in range(2, 1001):
    if not P[i]:
        continue
    for j in range(i * i, len(P), i):
        P[j] = 0
for i in range(2, len(P)):
    if P[i]:
        primes.append(i)

def solve():
    n = ii()
    r = 0
    for j in primes:
        if j > n:
            break
        c, m = n // j, n % j
        r += ((j - c % j) % j) * (m + 1)
        for i in range(1, c):
            r += j * ((j - i % j) % j)
        r %= mod
    c, m = n // 8, n % 8
    r = (r + c * 8 + max(0, m - 3) * 2) % mod
    print(r)

for _ in range(ii()):
    solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following aspects:

1. **Prime Number Handling**: The code uses a precomputed list of prime numbers up to 1000. It iterates over these primes and performs calculations. A test case that includes a number close to or just above 1000 might help expose issues with prime handling.

2. **Modulo Operation**: The code uses modulo operations, which can sometimes lead to incorrect results if not handled properly. A test case with a number that causes a lot of modulo operations might help.

3. **Edge Cases**: Numbers like 1, 2, and 1000 are edge cases that can help verify the correctness of the algorithm.

4. **Large Input**: Although the test input should be small-scale, a number close to the upper limit (10^6) can help ensure the algorithm handles large inputs correctly.

Given these considerations, a suitable test input might include a number that is just above 1000, a number that causes a lot of modulo operations, and a number close to the upper limit. Here is a small-scale test input:

```plaintext
4
1001
1002
1003
999999
```

This input includes numbers just above the prime limit (1000) and a number close to the upper limit (10^6), which should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Carousel of Combinations
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. The function $$$C(i,k)$$$ represents the number of distinct ways you can select $$$k$$$ distinct numbers from the set {$$$1, 2, \ldots, i$$$} and arrange them in a circle$$$^\dagger$$$.

Find the value of $$$$$$ \sum\limits_{i=1}^n \sum\limits_{j=1}^i \left( C(i,j) \bmod j \right). $$$$$$ Here, the operation $$$x \bmod y$$$ denotes the remainder from dividing $$$x$$$ by $$$y$$$.

Since this value can be very large, find it modulo $$$10^9+7$$$.

$$$^\dagger$$$ In a circular arrangement, sequences are considered identical if one can be rotated to match the other. For instance, $$$[1, 2, 3]$$$ and $$$[2, 3, 1]$$$ are equivalent in a circle.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$).

Output: For each test case, output a single integer on a new line — the value of the expression to be calculated modulo $$$10^9+7$$$.

Examples:
input:
4
1
3
6
314159
output:
0
4
24
78926217

Note: In the first test case, $$$C(1,1) \bmod 1 = 0$$$.

In the second test case:

- $$$C(1,1)=1$$$ (the arrangements are: $$$[1]$$$);
- $$$C(2,1)=2$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$);
- $$$C(2,2)=1$$$ (the arrangements are: $$$[1, 2]$$$);
- $$$C(3,1)=3$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$, $$$[3]$$$);
- $$$C(3,2)=3$$$ (the arrangements are: $$$[1, 2]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$);
- $$$C(3,3)=2$$$ (the arrangements are: $$$[1, 2, 3]$$$, $$$[1, 3, 2]$$$).

Buggy Code:
from bisect import bisect,bisect_left

from collections import *
from heapq import *
from math import gcd,ceil,sqrt,floor,inf,pi,lcm,isqrt,log2

from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *

#----------------------------------------------------------------------
import os
import sys

from io import BytesIO, IOBase
# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def RI():return list(map(int,list(input())))
def N(): return int(input())
def A(n,x=0):return [x]*n
def A2(n,m,x=0): return [[x]*m for i in range(n)]
def A3(a,b,c,x=0):return [[[x]*c for j in range(b)]for _ in range(a)]
def P2(a):
    for r in a:print(*r)
def G(n): return [[] for i in range(n)]
def c2(x):return x*(x-1)//2
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
def fmax(x,y):return y if y>x else x
def fmin(x,y):return y if y<x else x
def PS(a):return list(accumulate(a,initial=0))
#------------------------------------------------------------------------

from types import GeneratorType

def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc

mod=10**9+7
farr=[1]
ifa=[]

def fact(x,mod=0):
    if mod:
        while x>=len(farr):
            farr.append(farr[-1]*len(farr)%mod)
    else:
        while x>=len(farr):
            farr.append(farr[-1]*len(farr))
    return farr[x]

def ifact(x,mod):
    global ifa
    fact(x,mod)
    ifa.append(pow(farr[-1],mod-2,mod))
    for i in range(x,0,-1):
        ifa.append(ifa[-1]*i%mod)
    ifa.reverse()

def per(i,j,mod=0):
    if i<j: return 0
    if not mod:
        return fact(i)//fact(i-j)
    return farr[i]*ifa[i-j]%mod

def com(i,j,mod=0):
    if i<j: return 0
    if not mod:
        return per(i,j)//fact(j)
    return per(i,j,mod)*ifa[j]%mod

def catalan(n):
    return com(2*n,n)//(n+1)

def matrixmul(a,b,mod):
    return [[sum(x*y for x,y in zip(row,col))%mod for col in zip(*b)]for row in a]

def mpow(n,a,k,mod):
    mul=a
    res=None
    while k:
        if k&1:
            if res==None:
                res=mul
            else:
                res=matrixmul(res,mul,mod)
        k>>=1
        mul=matrixmul(mul,mul,mod)
    return res

def inverse(a,m):
    a%=m
    if a<=1: return a
    return ((1-inverse(m,a)*m)//a)%m

def exgcd(a,b):
    if b==0:
        return 1,0,a
    x,y,g=exgcd(b,a%b)
    x,y=y,(x-a//b*y)
    return x,y,g

def linearbase(a):
    res=[]
    for x in a:
        i=0
        for y in res:
            if x.bit_length()>y.bit_length():break
            if x.bit_length()==y.bit_length():
                x^=y
            i+=1
        if x:
            res.insert(i,x)
    return res

def lowbit(n):
    return n&-n
def GospersHack(k,n):
    cur=(1<<k)-1
    while cur<1<<n:
        yield cur
        lb=cur&-cur
        r=cur+lb
        cur=(r^cur)>>lb.bit_length()+1|r

def modui(ql):
    seq=sorted(range(q),key=lambda x:(ql[x][0]//B,ql[x][1]))
    res=[0]*q
    cnt=[0]*(max(a)+1)
    l,r=ql[seq[0]]
    cur=0
    for i in range(l,r+1):
        cur+=c2(cnt[a[i]])
        cnt[a[i]]+=1
    res[seq[0]]=cur
    #print(seq,ql,res,cnt)
    for i in range(1,q):
        li,ri=ql[seq[i]]
        if ri>r:
            for j in range(r+1,ri+1):
                cur+=c2(cnt[a[j]])
                cnt[a[j]]+=1
        else:
            for j in range(r,ri,-1):
                cnt[a[j]]-=1
                cur-=c2(cnt[a[j]])
        if li>l:
            for j in range(l,li):
                cnt[a[j]]-=1
                cur-=c2(cnt[a[j]])
        else:
            for j in range(l-1,li-1,-1):
                cur+=c2(cnt[a[j]])
                cnt[a[j]]+=1
        res[seq[i]]=cur
        l,r=li,ri
    return res
############################################data structure
class BIT:
    __slots__=('n','arr')
    def __init__(self,arr):
        self.arr=arr
        self.n=len(arr)

    def update(self,x,v):
        while x<self.n:
            self.arr[x]+=v
            x+=x&-x

    def query(self,x):
        ans=0
        while x:
            ans+=self.arr[x]
            x&=x-1
        return ans

    def display(self):
        print([self.query(i) for i in range(1,self.n)])

class ST:
    __slots__=('n','st','mx')
    def __init__(self,arr):#n!=0
        n=len(arr)
        mx=n.bit_length()#取不到
        self.st=[[0]*mx for i in range(n)]
        for i in range(n):
            self.st[i][0]=arr[i]
        for j in range(1,mx):
            for i in range(n-(1<<j)+1):
                self.st[i][j]=max(self.st[i][j-1],self.st[i+(1<<j-1)][j-1])
    def query(self,l,r):
        if l>r:return -inf
        s=(r+1-l).bit_length()-1
        return max(self.st[l][s],self.st[r-(1<<s)+1][s])

class TreeAncestor:
    def __init__(self, n, edges):
        mx=n.bit_length()
        self.par=[[-1]*n for _ in range(mx)]
        g = [[] for _ in range(n)]
        for u,v,w in edges:
            g[u].append((v,w))
            g[v].append((u,w))
        self.dep=[0]*n
        self.cnt=A2(n,27)
        @bootstrap
        def dfs(u,p):
            self.par[u]=p
            for v,w in g[u]:
                if v!=p:
                    cnt[v]=cnt[u].copy()
                    self.dep[v]=dep[u]+1
                    cnt[v][w]+=1
                    yield dfs(v,u)
            yield None
        dfs(0,-1)
        for i in range(mx-1):
            for u in range(n):
                if (p:=self.par[i][u])!=-1:
                    self.par[i+1][u]=self.par[i][p]

    def getKthAncestor(self,node, k):
        for i in range(k.bit_length()):
            if k>>i&1:
                node=self.par[i][node]
                if node<0:break
        return node

    def lca(self,u,v):
        if self.dep[u]<self.dep[v]:
            u,v=v,u
        u=self.getKthAncestor(u,self.dep[u]-self.dep[v])
        if u==v:return u
        for i in range(self.mx-1,-1,-1):
            if self.par[i][u]==self.par[i][v]:continue
            u=self.par[i][u]
            v=self.par[i][v]
        return self.par[0][u]
class DLN:
    def __init__(self,val):
        self.val=val
        self.pre=None
        self.next=None
class Trie:
    def __init__(self):
        self.end=0
        self.children={}

class DSU:#容量+路径压缩
    def __init__(self,n):
        self.c=[-1]*n

    def same(self,x,y):
        return self.find(x)==self.find(y)

    def find(self,x):
        if self.c[x]<0:
            return x
        self.c[x]=self.find(self.c[x])
        return self.c[x]

    def union(self,u,v):
        u,v=self.find(u),self.find(v)
        if u==v:
            return False
        if self.c[u]>self.c[v]:
            u,v=v,u
        self.c[u]+=self.c[v]
        self.c[v]=u
        return True

    def size(self,x): return -self.c[self.find(x)]
    def groups(self):
        n=len(self.c)
        result= [[] for _ in range(n)]
        for i in range(n):
            result[self.find(i)].append(i)
        return list(filter(lambda r: r, result))

class UF:#秩+路径+容量，边数
    def __init__(self,n):
        self.parent=[i for i in range(n)]
        self.ranks=[0]*n
        self.size=AI(n,1)
        self.edge=A(n)

    def find(self,x):
        if x!=self.parent[x]:
            self.parent[x]=self.find(self.parent[x])
        return self.parent[x]

    def union(self,u,v):
        pu,pv=self.find(u),self.find(v)
        if pu==pv:
            self.edge[pu]+=1
            return False
        if self.ranks[pu]>=self.ranks[pv]:
            self.parent[pv]=pu
            self.edge[pu]+=self.edge[pv]+1
            self.size[pu]+=self.size[pv]
            if self.ranks[pv]==self.ranks[pu]:
                self.ranks[pu]+=1
        else:
            self.parent[pu]=pv
            self.edge[pv]+=self.edge[pu]+1
            self.size[pv]+=self.size[pu]

#min,non-decreasing stack
def cartesian(a):
    ch=A2(n+1,2)
    stk=[0]
    for i in range(1,n+1):
        while stk and a[stk[-1]]>a[i]:
            stk.pop()
        ch[i][0]=ch[stk[-1]][1]
        ch[stk[-1]][1]=i
        stk.append(i)
    rt=stk[1]
    return rt,ch
#############################################################
def Prime(n):
    c=0
    prime=[]
    flag=[0]*(n+1)
    for i in range(2,n+1):
        if not flag[i]:
            prime.append(i)
            c+=1
        for j in range(c):
            if i*prime[j]>n: break
            flag[i*prime[j]]=prime[j]
            if i%prime[j]==0: break
    return flag

def rotate(self, a):#clockwise 90° return tuple
        return list(zip(*reversed(a)))

def lis(nums):
    res=[]
    for k in nums:
        i=bisect_left(res,k)
        if i==len(res):
            res.append(k)
        else:
            res[i]=k
    return len(res)

def RP(nums):#逆序对
    n = len(nums)
    s=set(nums)
    d={}
    for i,k in enumerate(sorted(s),1):
        d[k]=i
    bi=BIT([0]*(len(s)+1))
    ans=0
    for i in range(n-1,-1,-1):
        ans+=bi.query(d[nums[i]]-1)
        bi.update(d[nums[i]],1)
    return ans

def michange(a,b):
    d=defaultdict(deque)
    for i,x in enumerate(b):
        d[x].append(i)
    order=A(len(a))
    for i,x in enumerate(a):
        if not d:
            return -1
        order[i]=d[x].popleft()
    return RP(order)

###############################################graph
def nb(i,j,n,m):
    for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
        if 0<=ni<n and 0<=nj<m:
            yield ni,nj

def kru(n,e):
    dsu=DSU(n+1)
    e.sort()
    res=0
    cnt=n
    for w,u,v in e:
        if dsu.same(u,v):continue
        dsu.union(u,v)
        res+=w
        cnt-=1
    return res if cnt==1 else inf

def bell(s):#bellman-Ford
    dis=A(n,inf)
    dis[s]=0
    for i in range(n-1):
        for u,v,w in edge:
            if dis[v]>dis[u]+w:
                dis[v]=dis[u]+w
    change=A(n)
    for i in range(n):
        for u,v,w in edge:
            if dis[v]>dis[u]+w:
                dis[v]=dis[u]+w
                change[v]=1
    return dis

def dij(s,graph):
    d=[inf]*n
    d[s]=0
    heap=[(0,s)]
    while heap:
        dis,u=heappop(heap)
        if dis>d[u]:
            continue
        for v,w in graph[u]:
            if d[v]>d[u]+w:
                d[v]=d[u]+w
                heappush(heap,(d[v],v))
    return d

def topo(n):
    q=deque()
    res=[]
    for i in range(1,n+1):
        if ind[i]==0:
            q.append(i)
            res.append(i)
    while q:
        u=q.popleft()
        for v in g[u]:
            ind[v]-=1
            if ind[v]==0:
                q.append(v)
                res.append(v)
    return res

def diameter(n,g):
    vis=A(n)
    d=A(n)
    def dfs(u):
        vis[u]=1
        for v in g[u]:
            if not vis[v]:
                d[v]=d[u]+1
                dfs(v)
    dfs(0)
    ma=-1
    for i in range(n):
        if d[i]>ma:
            r=i
            ma=d[i]
    vis=A(n)
    d=A(n)
    dfs(r)
    return max(d)
#############################################################
def z_algorithm(s):
    n = len(s)
    if n == 0:return []
    z = [0]*n
    j = 0
    for i in range(1, n):
        z[i] = 0 if j + z[j] <= i else min(j + z[j] - i, z[i - j])
        while i + z[i] < n and s[z[i]] == s[i + z[i]]:
            z[i] += 1
        if j + z[j] < i + z[i]:
            j = i
    z[0] = n
    return z
def getnext(s):
    n=len(s)
    nxt=[-1]*(n+1)
    k=-1
    i=0
    while i<n:
        if k==-1 or s[i]==s[k]:
            nxt[i+1]=k+1
            i+=1
            k=nxt[i]
        else:
            k=nxt[k]
    #next val
    '''
    for i in range(1,n):
        if s[i]==s[nxt[i]]:
            nxt[i]=nxt[nxt[i]]'''
    return nxt

def kmp(s1,s2,nxt,i=0):
    #nxt=getnext(s2)
    l1,l2=len(s1),len(s2)
    j=0
    while i<l1 and j<l2:
        if j==-1 or s1[i]==s2[j]:
            i+=1
            j+=1
        else:
            j=nxt[j]
    return j
    '''if j==l2:
        return i-j
    return -1'''

@bootstrap
def dfs(u,p):
    for v in g[u]:
        if v!=p:
            yield dfs(v,u)
    yield None

from random import randint,shuffle
def ra(n,a,b):
    return [randint(a,b) for i in range(n)]

#mod=998244353
flag=Prime(10**6)
t=N()
for i in range(t):
    n=N()
    ans=0
    for j in range(2,n+1):
        if flag[j]==0:
            for i in range(j,n+1,j):
                ans+=-(i//j)%j*j
            ans-=-(n//j)%j*(i+j-1-n)
            ans%=mod
    for i in range(4,n+1):
        if i//2%4==2 or i//2%4==3:
            ans+=2
    print(ans)

'''
sys.setrecursionlimit(200000)
sys.stdout.flush()
import threading
threading.sta1ck_size(10**8)
t=threading.Thr
ead(target=main)
t.start()
t.join()

'''

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

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

Think step by step.

Response:
```plaintext
4
1
3
6
314159
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Carousel of Combinations
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. The function $$$C(i,k)$$$ represents the number of distinct ways you can select $$$k$$$ distinct numbers from the set {$$$1, 2, \ldots, i$$$} and arrange them in a circle$$$^\dagger$$$.

Find the value of $$$$$$ \sum\limits_{i=1}^n \sum\limits_{j=1}^i \left( C(i,j) \bmod j \right). $$$$$$ Here, the operation $$$x \bmod y$$$ denotes the remainder from dividing $$$x$$$ by $$$y$$$.

Since this value can be very large, find it modulo $$$10^9+7$$$.

$$$^\dagger$$$ In a circular arrangement, sequences are considered identical if one can be rotated to match the other. For instance, $$$[1, 2, 3]$$$ and $$$[2, 3, 1]$$$ are equivalent in a circle.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$).

Output: For each test case, output a single integer on a new line — the value of the expression to be calculated modulo $$$10^9+7$$$.

Examples:
input:
4
1
3
6
314159
output:
0
4
24
78926217

Note: In the first test case, $$$C(1,1) \bmod 1 = 0$$$.

In the second test case:

- $$$C(1,1)=1$$$ (the arrangements are: $$$[1]$$$);
- $$$C(2,1)=2$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$);
- $$$C(2,2)=1$$$ (the arrangements are: $$$[1, 2]$$$);
- $$$C(3,1)=3$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$, $$$[3]$$$);
- $$$C(3,2)=3$$$ (the arrangements are: $$$[1, 2]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$);
- $$$C(3,3)=2$$$ (the arrangements are: $$$[1, 2, 3]$$$, $$$[1, 3, 2]$$$).

Buggy Code:
from sys import stdin
input = stdin.readline
mod = 10**9+7; ANS = []
sieve = [min(i,2) if i%2==0 else i for i in range(10**6+1)]
prime = [2]
for i in range(3,1001):
    if sieve[i] == i:
        for j in range(2*i,10**6+1,i): sieve[j] = i
for i in range(3,10**6+1,2):
    if i == sieve[i]: prime.append(i)
def suma(num,p):
    if p == 4:
        return (2*min(4,num%8) + 8*(num//8))%mod
    else:
        res,rres = (num%(p**2))//p,num%p
        ans = ((p*p*(p-1))//2)*(num//(p**2))%mod+p*((res*(2*p-1-res))//2)+(p-1-res)*rres
        return ans%mod
def solve(N):
    if N == 1: ANS.append('0'); return
    ans = 0 if N<4 else suma(N-3,4)
    for i in range(78498):
        p = prime[i]
        if p > N: ANS.append(str(ans)); break
        ans += suma(N-p+1,p); ans %= mod
    else: ANS.append(str(ans))
for _ in range(int(input())): solve(int(input()))
print('\n'.join(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
4
1
3
10
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Carousel of Combinations
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. The function $$$C(i,k)$$$ represents the number of distinct ways you can select $$$k$$$ distinct numbers from the set {$$$1, 2, \ldots, i$$$} and arrange them in a circle$$$^\dagger$$$.

Find the value of $$$$$$ \sum\limits_{i=1}^n \sum\limits_{j=1}^i \left( C(i,j) \bmod j \right). $$$$$$ Here, the operation $$$x \bmod y$$$ denotes the remainder from dividing $$$x$$$ by $$$y$$$.

Since this value can be very large, find it modulo $$$10^9+7$$$.

$$$^\dagger$$$ In a circular arrangement, sequences are considered identical if one can be rotated to match the other. For instance, $$$[1, 2, 3]$$$ and $$$[2, 3, 1]$$$ are equivalent in a circle.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$).

Output: For each test case, output a single integer on a new line — the value of the expression to be calculated modulo $$$10^9+7$$$.

Examples:
input:
4
1
3
6
314159
output:
0
4
24
78926217

Note: In the first test case, $$$C(1,1) \bmod 1 = 0$$$.

In the second test case:

- $$$C(1,1)=1$$$ (the arrangements are: $$$[1]$$$);
- $$$C(2,1)=2$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$);
- $$$C(2,2)=1$$$ (the arrangements are: $$$[1, 2]$$$);
- $$$C(3,1)=3$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$, $$$[3]$$$);
- $$$C(3,2)=3$$$ (the arrangements are: $$$[1, 2]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$);
- $$$C(3,3)=2$$$ (the arrangements are: $$$[1, 2, 3]$$$, $$$[1, 3, 2]$$$).

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

public class Main {
    static boolean[] isPrime;
    private static void solve(MyScanner in, PrintWriter out) {
        if (isPrime == null) build();
        int n = in.nextInt();
        long res = (n + 1) / 8 * 8 + ((n + 1) % 8 >= 5 ? (n + 1) % 4 * 2 : 0);
        int MOD = 1_000_000_007;
        for (int j = 2; j <= n; j++) {
            if (!isPrime[j]) continue;
            res = (res + (long) j * (j - 1) / 2 % MOD * j % MOD * ((n + 1) / j / j)) % MOD;
            long v = ((n + 1) / j % j);
            if (v >= 2) res = (res + ((v - 1) * (j + j - v) / 2 % MOD * j % MOD)) % MOD;
            if (v >= 1) res = (res + ((j - v) % MOD * ((n + 1) % j))) % MOD;
        }
        out.println(res);
    }

    private static void build() {
        isPrime = new boolean[1000001];
        Arrays.fill(isPrime, true);
        for (int i = 2; i <= 1000000; i++) {
            if (isPrime[i]) {
                for (int j = i + i; j <= 1000000; j += i) isPrime[j] = false;
            }
        }
    }

    public static void main(String[] args) {
        MyScanner in = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        for (int t = in.nextInt(); t > 0; t--) solve(in, out);
        out.close();
    }

    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    }
    //--------------------------------------------------------
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Carousel of Combinations
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given an integer $$$n$$$. The function $$$C(i,k)$$$ represents the number of distinct ways you can select $$$k$$$ distinct numbers from the set {$$$1, 2, \ldots, i$$$} and arrange them in a circle$$$^\dagger$$$.

Find the value of $$$$$$ \sum\limits_{i=1}^n \sum\limits_{j=1}^i \left( C(i,j) \bmod j \right). $$$$$$ Here, the operation $$$x \bmod y$$$ denotes the remainder from dividing $$$x$$$ by $$$y$$$.

Since this value can be very large, find it modulo $$$10^9+7$$$.

$$$^\dagger$$$ In a circular arrangement, sequences are considered identical if one can be rotated to match the other. For instance, $$$[1, 2, 3]$$$ and $$$[2, 3, 1]$$$ are equivalent in a circle.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$).

Output: For each test case, output a single integer on a new line — the value of the expression to be calculated modulo $$$10^9+7$$$.

Examples:
input:
4
1
3
6
314159
output:
0
4
24
78926217

Note: In the first test case, $$$C(1,1) \bmod 1 = 0$$$.

In the second test case:

- $$$C(1,1)=1$$$ (the arrangements are: $$$[1]$$$);
- $$$C(2,1)=2$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$);
- $$$C(2,2)=1$$$ (the arrangements are: $$$[1, 2]$$$);
- $$$C(3,1)=3$$$ (the arrangements are: $$$[1]$$$, $$$[2]$$$, $$$[3]$$$);
- $$$C(3,2)=3$$$ (the arrangements are: $$$[1, 2]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$);
- $$$C(3,3)=2$$$ (the arrangements are: $$$[1, 2, 3]$$$, $$$[1, 3, 2]$$$).

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

public class Main {
    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        Task solver = new Task();
        solver.solve(in.nextInt(), in, out);
//        solver.duiPai();
        out.close();
    }

    static class Task {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int[] arr = new int[testNumber];
            for (int t = 0; t < testNumber; t++) {
                arr[t] = in.nextInt();
            }
            int max = Arrays.stream(arr).max().orElse(0);
            Prime prime = new Prime(max + 5);
            Factorial fac = new Factorial(max + 5, (int) (1e9 + 7));
            for (int t = 0; t < testNumber; t++) {
                out.println(new Solution(fac, prime).solve(arr[t]));
            }
        }

        public void duiPai() {
            Random random = new Random();
            for (int t = 0; t < 100; t++) {
                int n = random.nextInt(10) + 5;
                int[] a = genArr(n, 1, 9);
                System.out.println(Arrays.toString(a));
            }
        }

        public int[] genArr(int n, int l, int r) {
            int[] arr = new int[n];
            Random random = new Random();
            for (int i = 0; i < n; i++) {
                arr[i] = random.nextInt(r - l + 1) + l;
            }
            return arr;
        }
    }

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

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

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

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

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

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

class Solution {
    long mod = (long) (1e9 + 7);
    Factorial fac;
    Prime prime;

    public Solution(Factorial fac, Prime prime) {
        this.fac = fac;
        this.prime = prime;
    }

    public long solve(int n) {
        long ans = 0;
        for (int j = 2; j <= n; j++) {
            long t = n - j + 1;
            if (j == 4) {
                long c = t / 8;
                ans += c * 8;
                long m = t % 8;
                if (m < 4) {
                    ans += m * 2;
                } else {
                    ans += 8;
                }
                ans %= mod;
                continue;
            }
            if (!prime.isPrime(j)) {
                continue;
            }
            if (t >= (long) j * j) {
                long c1 = t / ((long) j * j);
                ans += (long) j * (j - 1) / 2 % mod * j % mod * c1 % mod;
                ans %= mod;
                t %= ((long) j * j);
            }
            long l = t / j;
            long m = t % j;
            long a0 = j - l;
            ans += (a0 + (j - 1)) * l / 2 % mod * j % mod;
            ans %= mod;
            ans += m * (j - l - 1) % mod;
            ans %= mod;
        }
        return (ans % mod + mod) % mod;
    }
}

class Factorial {
    long[] fac;
    long[] inv;
    long mod;

    public Factorial(long[] fac, long[] inv, long mod) {
        this.mod = mod;
        this.fac = fac;
        this.inv = inv;
        fac[0] = inv[0] = 1;
        int n = fac.length;
        for (int i = 1; i < n; i++) {
            fac[i] = i;
            fac[i] = (int) (fac[i] * fac[i - 1] % mod);
        }
        inv[n - 1] = BigInteger.valueOf(fac[n - 1]).modInverse(BigInteger.valueOf(mod)).intValue();
        for (int i = n - 2; i >= 1; i--) {
            inv[i] = (int) (inv[i + 1] * (i + 1) % mod);
        }
    }

    public Factorial(int limit, int mod) {
        this(new long[Math.min(limit + 1, mod)], new long[Math.min(limit + 1, mod)], mod);
    }

    public long getMod() {
        return mod;
    }

    public long fac(int n) {
        if (n >= mod) {
            return 0;
        }
        return fac[n];
    }

    public long inv(int n) {
        if (n >= mod) {
            throw new IllegalArgumentException();
        }
        return inv[n];
    }

    public long comb(int n, int m) {
        if (n < m) {
            return 0;
        }
        return fac[n] * inv[m] % mod * inv[n - m] % mod;
    }

    public long combInv(int n, int m) {
        if (n < m) {
            throw new IllegalArgumentException();
        }
        return inv[n] * fac[m] % mod * fac[n - m] % mod;
    }
}

class Prime {
    /**
     * 记录每个数的最小质因子，当最小质因子等于该数时，说明是质数
     */
    int[] v;

    /**
     * 质数列表
     */
    List<Integer> primeList = new ArrayList<>();

    public Prime(int n) {
        v = primes(n);
        for (int i = 2; i < v.length; i++) {
            if (v[i] == i) {
                primeList.add(i);
            }
        }
    }

    public boolean isPrime(int n) {
        return n >= 2 && v[n] == n;
    }

    /**
     * 质因数分解，key为质数，value为该质因数的个数，时间复杂度O(log(n))
     */
    public Map<Integer, Integer> divide(int n) {
        Map<Integer, Integer> map = new HashMap<>();
        while (n > 1) {
            map.put(v[n], map.getOrDefault(v[n], 0) + 1);
            n /= v[n];
        }
        return map;
    }

    /**
     * 埃氏筛，时间复杂度O(n * log(log(n)))
     */
    public static int[] primes(int n) {
        int[] v = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            v[i] = i;
        }
        for (int i = 2; i * i <= n; i++) {
            if (v[i] != i) {
                continue;
            }
            for (int j = i * i; j <= n; j += i) {
                if (v[j] == j) {
                    v[j] = i;
                }
            }
        }
        return v;
    }
}

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

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

Think step by step.

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

1. **Small Values of `n`**: Test with small values of `n` to ensure the basic logic is correct.
2. **Prime Values of `n`**: Test with prime values of `n` to ensure the prime number handling is correct.
3. **Composite Values of `n`**: Test with composite values of `n` to ensure the composite number handling is correct.
4. **Boundary Values**: Test with boundary values of `n` to ensure the limits are handled correctly.

Given the constraints and the nature of the problem, a small-scale test input with a few test cases that cover these scenarios should be sufficient. Here is a test input that includes these cases:

```plaintext
4
1
4
7
12
```

### Explanation:
1. **Test Case 1: `n = 1`**
   - This is the smallest possible value of `n`. It should return `0` because there are no valid combinations for `n = 1`.

2. **Test Case 2: `n = 4`**
   - This is a small composite number. It should test the handling of composite numbers and the special case for `j = 4`.

3. **Test Case 3: `n = 7`**
   - This is a small prime number. It should test the handling of prime numbers and the prime number logic.

4. **Test Case 4: `n = 12`**
   - This is a slightly larger composite number. It should test the handling of larger composite numbers and the general logic.

This test input should help expose any issues in the buggy code, particularly in the handling of prime and composite numbers, as well as the special case for `j = 4`.