**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

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

    for n, q, matrix, queries in test_cases:
        for (x1, y1, x2, y2) in queries:
            flatten_list = []
            for i in range(x1 - 1, x2):  # converting to 0-indexed for Python access
                for j in range(y1 - 1, y2):
                    flatten_list.append(matrix[i][j])

            weighted_sum = sum(value * (idx + 1) for idx, value in enumerate(flatten_list))
            results.append(str(weighted_sum))

    return ' '.join(results)

# Example of input processing and calling the function
if __name__ == "__main__":
    import sys
    input = sys.stdin.read
    data = input().splitlines()

    index = 0
    t = int(data[index])  # number of test cases
    index += 1
    test_cases = []

    for _ in range(t):
        n, q = map(int, data[index].split())
        index += 1
        matrix = []
        for _ in range(n):
            row = list(map(int, data[index].split()))
            matrix.append(row)
            index += 1

        queries = []
        for _ in range(q):
            x1, y1, x2, y2 = map(int, data[index].split())
            queries.append((x1, y1, x2, y2))
            index += 1

        test_cases.append((n, q, matrix, queries))

    results = process_pancake_factory(t, test_cases)
    print(results)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to create such a test input:

1. **Edge Case with Minimum and Maximum Values**: Test with the smallest and largest possible values for \( n \) and \( q \).
2. **Edge Case with Submatrix Boundaries**: Test with queries that include the entire matrix, single elements, and submatrices of various sizes.
3. **Edge Case with Repeated Elements**: Test with matrices that have repeated elements to ensure the weighted sum calculation is correct.
4. **Edge Case with Large Elements**: Test with matrices that have large elements to ensure the weighted sum does not overflow.

Given the constraints and the need for a small-scale test input, let's create a test case with \( n = 4 \) and \( q = 3 \) that includes a variety of submatrices and large elements.

```plaintext
1
4 3
1000000 1000000 1000000 1000000
1000000 1000000 1000000 1000000
1000000 1000000 1000000 1000000
1000000 1000000 1000000 1000000
1 1 4 4
2 2 3 3
1 2 4 3
```

### Explanation:
- **Test Case 1**:
  - **Matrix Size**: \( n = 4 \)
  - **Number of Queries**: \( q = 3 \)
  - **Matrix**: A 4x4 matrix filled with the maximum value \( 1000000 \)
  - **Queries**:
    - **Query 1**: Flattens the entire matrix (1, 1, 4, 4)
    - **Query 2**: Flattens a 2x2 submatrix in the middle (2, 2, 3, 3)
    - **Query 3**: Flattens a submatrix from the second column to the third column (1, 2, 4, 3)

This test case is designed to:
- **Test the entire matrix**: Ensures that the code can handle a large number of elements.
- **Test a smaller submatrix**: Ensures that the code can handle submatrices of different sizes.
- **Test edge cases**: Ensures that the code can handle the boundaries of the matrix correctly.
- **Test large values**: Ensures that the weighted sum calculation does not overflow.

By using this test input, we can effectively expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            int n = sc.nextInt();
            int q = sc.nextInt();
            long[][] pA = new long[n + 1][n + 1];
            long[][] pB = new long[n + 1][n + 1];
            long[][] pC = new long[n + 1][n + 1];
            for (int i = 1; i <= n; i++){
                for (int j = 1; j <= n; j++){
                    int v = sc.nextInt();
                    pA[i][j] = pA[i - 1][j] + pA[i][j - 1] - pA[i - 1][j - 1] + v;
                    pB[i][j] = pB[i - 1][j] + pB[i][j - 1] - pB[i- 1][j - 1] + (long)i * v;
                    pC[i][j] = pC[i - 1][j] + pC[i][j - 1] - pC[i - 1][j - 1] + (long)j * v;
                }
            }
            for (int k = 0; k < q; k++){
                int x1 = sc.nextInt();
                int y1 = sc.nextInt();
                int x2 = sc.nextInt();
                int y2 = sc.nextInt();
                long c = (long)(y2 - y1 + 1);
                long sS = pA[x2][y2] - pA[x1 - 1][y2] - pA[x2][y1 - 1] + pA[x1 - 1][y1 - 1];
                long sI = pB[x2][y2] - pB[x1 - 1][y2] - pB[x2][y1 - 1] + pB[x1 - 1][y1 - 1];
                long sJ = pC[x2][y2] - pC[x1 - 1][y2] - pC[x2][y1 - 1] + pC[x1 - 1][y1 - 1];
                long ans = c * sI + sJ + (1L - c * x1 - y1) * sS;
                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
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

Buggy Code:
import java.util.Scanner;

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

        int t = scanner.nextInt();

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

            long[][] matrix = new long[n][n];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    matrix[i][j] = scanner.nextLong();
                }
            }

            // Compute prefix sums
            long[][] prefixSum = new long[n + 1][n + 1];
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    prefixSum[i][j] = prefixSum[i - 1][j] + prefixSum[i][j - 1] - prefixSum[i - 1][j - 1] + matrix[i - 1][j - 1];
                }
            }

            // Process each query
            for (int qi = 0; qi < q; qi++) {
                int x1 = scanner.nextInt() - 1;
                int y1 = scanner.nextInt() - 1;
                int x2 = scanner.nextInt() - 1;
                int y2 = scanner.nextInt() - 1;

                // Calculate the sum of the submatrix
                long submatrixSum = prefixSum[x2 + 1][y2 + 1] - prefixSum[x1][y2 + 1] - prefixSum[x2 + 1][y1] + prefixSum[x1][y1];

                // Flatten the submatrix and compute the required sum
                long result = 0;
                for (int i = 0; i <= x2 - x1; i++) {
                    for (int j = 0; j <= y2 - y1; j++) {
                        result += (i * (y2 - y1 + 1) + j + 1) * matrix[x1 + i][y1 + j];
                    }
                }

                System.out.print(result + " ");
            }
            System.out.println();
        }

        scanner.close();
    }
}

// XBUMhTsEK45kLbJuReS

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The buggy code uses a prefix sum array to calculate the sum of the submatrix but then manually flattens the submatrix to compute the required sum. The manual flattening process is where the bug might be introduced.

### Key Points to Consider:
1. **Edge Cases:**
   - Queries that cover the entire matrix.
   - Queries that cover a single cell.
   - Queries that cover a row or a column.
2. **Large Values:**
   - Matrices with large values to test the sum calculation.
3. **Corner Cases:**
   - Queries where the submatrix is a single cell.
   - Queries where the submatrix is a row or a column.
   - Queries where the submatrix is the entire matrix.

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

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

### Explanation:
1. **First Test Case:**
   - Matrix size: 4x4
   - Queries:
     - Query 1: (1, 1) to (4, 4) (entire matrix)
     - Query 2: (2, 2) to (3, 3) (a 2x2 submatrix)
     - Query 3: (1, 2) to (4, 3) (a 4x2 submatrix)

2. **Second Test Case:**
   - Matrix size: 3x3
   - Queries:
     - Query 1: (1, 1) to (1, 3) (a row)
     - Query 2: (1, 3) to (3, 3) (a column)
     - Query 3: (2, 2) to (2, 2) (a single cell)

This test input covers a variety of scenarios, including edge cases and large values, which 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: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define vvi vector<vector<ll>>
#define vi vector<ll>
#define get(n) int n; cin >> n;
using namespace std;

vvi M(2001, vi(2001));
vvi Sum(2001, vi(2001));
vvi x_Sum(2001, vi(2001));
vvi y_Sum(2001, vi(2001));
ll n, x1, y_start, x2, y2;
int q;
vi dp_x(2001);
vi dp_y(2001);
vi dp_sum(2001);

ll return_sum_x(){
    ll ans = 0;
    ans += x_Sum[x2][y2];
    ans -= x_Sum[x2][y_start-1];
    ans -= x_Sum[x1-1][y2];
    ans += x_Sum[x1-1][y_start-1];
    return ans;
}

ll return_sum_y(){
    ll ans = 0;
    ans += y_Sum[x2][y2];
    ans -= y_Sum[x2][y_start-1];
    ans -= y_Sum[x1-1][y2];
    ans += y_Sum[x1-1][y_start-1];
    return ans;
}

ll return_sum_normal(){
    ll ans = 0;
    ans += Sum[x2][y2];
    ans -= Sum[x2][y_start-1];
    ans -= Sum[x1-1][y2];
    ans += Sum[x1-1][y_start-1];
    return ans;
}

void cout_vec(vvi &v){
    for(int i = 1; i<= n ; i++){
        for(int j = 1; j <= n ; j++){
            cout << v[i][j] << " ";
        }
        cout << "\n";
    }
    cout <<"\nCOMP\n\n";
}

void solve(){
    ll n;
    int q;

    cin >> n >> q;
    for(int i = 1; i <= n ; i++){
        for(int j =1 ; j <= n ; j++){
            cin >> M[i][j];
        }
    }
    for(ll i = 1 ; i<= n ; i++){
        for(ll j = 1 ; j <=n ; j++){
            Sum[i][j] = Sum[i-1][j] + dp_sum[j-1] + M[i][j];
            dp_sum[j] = dp_sum[j-1] + M[i][j];
            x_Sum[i][j] = x_Sum[i-1][j] + dp_x[j-1] + i*M[i][j];
            dp_x[j] = dp_x[j-1] + (i*M[i][j]);
            y_Sum[i][j] = y_Sum[i-1][j] + dp_y[j-1] + j*M[i][j];
            dp_y[j] = dp_y[j-1] + (j*M[i][j]);
        }
    }
    while(q--){
        cin >> x1 >> y_start >> x2 >> y2;
        printf("%d ", (return_sum_x())*(y2-y_start+1) + return_sum_y() - (return_sum_normal())*((y2-y_start+1)*x1 + y_start - 1));
    }

    printf("\n");
}

int main(){
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    get(testcases);
    while(testcases--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
1 1 1 1
1 1 3 3
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

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

int main()
{
    int tt;
    cin >> tt;
    while (tt > 0) {
        int n, q;
        cin >> n >> q;
        vector<vector<int>> v(n, vector<int>(n));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cin >> v[i][j];
            }
        }
        for (int z = 1; z <= q; z++) {
            int x1, y1, x2, y2;
            cin >> x1 >> y1 >> x2 >> y2;
            int cnt = 1; // Reset cnt for every query
            int sum = 0;
            for (int k = x1 - 1; k < x2; k++) {
                for (int l = y1 - 1; l < y2; l++) {
                    sum += cnt * v[k][l];
                    cnt++; // Increment weight
                }
            }
            cout << sum << " ";
        }
        cout << endl;
        tt--;
    }
    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 buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Matrix Size and Query Complexity**: The code processes an \( n \times n \) matrix and handles up to \( q \) queries. We should consider a small but non-trivial size for \( n \) and \( q \) to ensure the test case is manageable yet challenging.

2. **Edge Cases**:
   - **Single Element Query**: A query that selects a single element.
   - **Full Matrix Query**: A query that selects the entire matrix.
   - **Row and Column Queries**: Queries that select entire rows or columns.
   - **Overlapping Queries**: Queries that overlap in different ways to test the accumulation of the sum.

3. **Large Values**: Use large values in the matrix to ensure the sum calculations are significant and can reveal overflow or precision issues.

4. **Indexing**: Ensure that the input values for \( x1, y1, x2, y2 \) are 1-based as specified in the problem, and the code should handle these correctly.

Given these considerations, let's construct a test input:

```plaintext
3
4 4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
1 1 1 1
1 1 4 4
2 2 2 2
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 3 3
2 2 3 3
1 1 1 1
2 2
1 2
3 4
1 1 2 2
1 2 2 1
```

### Explanation:
1. **Test Case 1**:
   - **Matrix Size**: \( 4 \times 4 \)
   - **Queries**:
     - **Query 1**: Single element (1,1) -> Should return \( 1 \)
     - **Query 2**: Full matrix (1,1) to (4,4) -> Should return \( 500 \)
     - **Query 3**: Single element (2,2) -> Should return \( 12 \)
     - **Query 4**: Submatrix (1,2) to (4,3) -> Should return \( 168 \)

2. **Test Case 2**:
   - **Matrix Size**: \( 3 \times 3 \)
   - **Queries**:
     - **Query 1**: Full matrix (1,1) to (3,3) -> Should return \( 14 \)
     - **Query 2**: Submatrix (2,2) to (3,3) -> Should return \( 42 \)
     - **Query 3**: Single element (1,1) -> Should return \( 1 \)

3. **Test Case 3**:
   - **Matrix Size**: \( 2 \times 2 \)
   - **Queries**:
     - **Query 1**: Full matrix (1,1) to (2,2) -> Should return \( 10 \)
     - **Query 2**: Submatrix (1,2) to (2,1) -> Should return \( 8 \)

This test input covers a variety of scenarios, including single elements, full matrices, and submatrices, and uses a small matrix size to ensure it is manageable while still being challenging.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

Buggy Code:
t = int(input())
results = []
for _ in range(t):
    n, q = map(int, input().split())
    M = [list(map(int, input().split())) for _ in range(n)]
    prefix_sum = [[0] * (n + 1) for _ in range(n + 1)]
    for i in range(1, n + 1):
        for j in range(1, n + 1):
            prefix_sum[i][j] = M[i-1][j-1] + prefix_sum[i-1][j] + prefix_sum[i][j-1] - prefix_sum[i-1][j-1]
    case_results = []
    for __ in range(q):
        x1, y1, x2, y2 = map(int, input().split())
        x1 -= 1; y1 -= 1; x2 -= 1; y2 -= 1
        weighted_sum = 0
        index = 1
        for i in range(x1, x2 + 1):
            for j in range(y1, y2 + 1):
                weighted_sum += M[i][j] * index
                index += 1
        case_results.append(weighted_sum)
    results.append(" ".join(map(str, case_results)))
print("\n".join(results))

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

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

Think step by step.

Response:
```plaintext
1
3 4
1 2 3
4 5 6
7 8 9
1 1 1 1
1 1 3 3
2 2 3 3
1 2 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

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

    M = [[0] * (n+1) for _ in range(n+1)]
    for r in range(1, n+1):
        row = list(map(int, input().split()))
        for c in range(1, n+1):
            M[r][c] = row[c-1]

    S = [[0] * (n+1) for _ in range(n+1)]
    Sr = [[0] * (n+1) for _ in range(n+1)]
    Sc = [[0] * (n+1) for _ in range(n+1)]

    for r in range(1, n+1):
            for c in range(1, n+1):
                S[r][c] = S[r-1][c] + S[r][c-1] - S[r-1][c-1] + M[r][c]
                Sr[r][c] = Sr[r-1][c] + Sr[r][c-1] - Sr[r-1][c-1] + r * M[r][c]
                Sc[r][c] = Sc[r-1][c] + Sc[r][c-1] - Sc[r-1][c-1] + c * M[r][c]

    output = []
    for _ in range(q):
        x1, y1, x2, y2 = map(int, input().split())
        width = y2 - y1 + 1

        sum_S = S[x2][y2] - S[x1-1][y2] - S[x2][y1-1] + S[x1-1][y1-1]
        sum_Sr = Sr[x2][y2] - Sr[x1-1][y2] - Sr[x2][y1-1] + Sr[x1-1][y1-1]
        sum_Sc = Sc[x2][y2] - Sc[x1-1][y2] - Sc[x2][y1-1] + Sc[x1-1][y1-1]

        term1 = width * (sum_Sr - x1 * sum_S)
        term2 = sum_Sc - (y1 - 1) * sum_S
        total = term1 + term2
        output.append(str(total))

    print(" ".join(output))

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

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

Think step by step.

Response:
```plaintext
1
4 4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
1 1 1 1
1 1 2 2
1 1 3 3
1 1 4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

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

public class Main{

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);

        int t = Integer.parseInt(br.readLine());
        while (t-- > 0) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int q = Integer.parseInt(st.nextToken());

            int[][] matrix = new int[n][n];
            for (int i = 0; i < n; i++) {
                st = new StringTokenizer(br.readLine());
                for (int j = 0; j < n; j++) {
                    matrix[i][j] = Integer.parseInt(st.nextToken());
                }
            }

            for (int i = 0; i < q; i++) {
                st = new StringTokenizer(br.readLine());
                int x1 = Integer.parseInt(st.nextToken()) - 1; // 0-based indexing
                int y1 = Integer.parseInt(st.nextToken()) - 1;
                int x2 = Integer.parseInt(st.nextToken()) - 1;
                int y2 = Integer.parseInt(st.nextToken()) - 1;

                long sum = 0;
                int pos = 1;
                for (int x = x1; x <= x2; x++) {
                    for (int y = y1; y <= y2; y++) {
                        sum += matrix[x][y] * pos++;
                    }
                }
                pw.print(sum + " ");
            }
            pw.println();
        }
        pw.flush();
    }
}

// UwGzOeZJ9QrtWVXS1g7

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

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

Think step by step.

Response:
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
1 1 1 1
1 1 3 3
1 1 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

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

public class Main{
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        StringTokenizer st = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(st.nextToken());
        for (int test = 0; test < t; test++) {
            st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int q = Integer.parseInt(st.nextToken());
            int[][] M = new int[n + 1][n + 1];
            for (int i = 1; i <= n; i++) {
                st = new StringTokenizer(br.readLine());
                for (int j = 1; j <= n; j++) {
                    M[i][j] = Integer.parseInt(st.nextToken());
                }
            }
            int[][] prefix_sum = new int[n + 1][n + 1];
            int[][] prefix_sum_j = new int[n + 1][n + 1];
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    prefix_sum[i][j] = prefix_sum[i][j - 1] + M[i][j];
                    prefix_sum_j[i][j] = prefix_sum_j[i][j - 1] + M[i][j] * j;
                }
            }
            for (int query = 0; query < q; query++) {
                st = new StringTokenizer(br.readLine());
                int x1 = Integer.parseInt(st.nextToken());
                int y1 = Integer.parseInt(st.nextToken());
                int x2 = Integer.parseInt(st.nextToken());
                int y2 = Integer.parseInt(st.nextToken());
                int sum1 = 0;
                int sum2 = 0;
                for (int i = x1; i <= x2; i++) {
                    int row_sum = prefix_sum[i][y2] - prefix_sum[i][y1 - 1];
                    sum1 += (i - x1) * row_sum;
                    int sum_j = prefix_sum_j[i][y2] - prefix_sum_j[i][y1 - 1];
                    sum2 += sum_j - (y1 - 1) * row_sum;
                }
                int total_sum = (y2 - y1 + 1) * sum1 + sum2;
                sb.append(total_sum).append(" ");
            }
            sb.append("\n");
        }
        System.out.print(sb.toString());
    }
}

// somlLeDvfT6GMIAXzWU

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

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

Think step by step.

Response:
```plaintext
1
4 4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
1 1 4 4
1 1 2 2
2 2 3 3
1 3 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define po pop_back
#define vl vector<long long>
#define vi vector<int>
#define vt vector<int64_t>
#define vs vector<string>
#define vii vector<pair<int, int>>
#define vll vector<pair<long long, long long>>
#define si set<int>
#define mii map<int,int>
#define umii unordered_map<int, int>
#define umll unordered_map<long long, long long>
#define all(v) sort(v.begin(),v.end())
#define rall(v) sort(v.rbegin(),v.rend())
#define sortall(a) sort(a,a+n)
#define string(s) sort(s.begin(), s.end())
#define rstring(s) reverse(s.begin(), s.end())
#define repa(a,n) for (ll i = a; i < (ll)(n); i++)
#define repb(b,n) for (ll i = b; i <= (ll)(n); i++)
#define PI 3.1415926535897932384626433832795
#define print(x) cout<<(x)<<endl;

int main() {
#ifndef ONLINE_JUDGE
freopen("inputf.in", "r", stdin);
freopen("outputf.in", "w", stdout);
#endif

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

        vector<vector<ll>> M(n, vector<ll>(n, 0));
        for(ll r = 0; r < n; r++){
            for(ll c = 0; c < n; c++){
                cin >> M[r][c];
            }
        }

        vector<vector<ll>> S(n+1, vector<ll>(n+1, 0));
        vector<vector<ll>> Sr(n+1, vector<ll>(n+1, 0));
        vector<vector<ll>> Sc(n+1, vector<ll>(n+1, 0));

        for(ll r = 1; r <= n; r++){
            ll current_S = 0;
            ll current_Sr = 0;
            ll current_Sc = 0;
            for(ll c = 1; c <= n; c++){
                ll val = M[r-1][c-1];
                current_S += val;
                current_Sr += val * r;
                current_Sc += val * c;

                S[r][c] = S[r-1][c] + current_S;
                Sr[r][c] = Sr[r-1][c] + current_Sr;
                Sc[r][c] = Sc[r-1][c] + current_Sc;
            }
        }

        string s;
        s.reserve(q * 20);
        for(ll i = 0; i < q; i++){
            ll x1, y1, x2, y2;
            cin >> x1 >> y1 >> x2 >> y2;

            ll S_sub = S[x2][y2] - S[x1-1][y2] - S[x2][y1-1] + S[x1-1][y1-1];
            ll Sr_sub = Sr[x2][y2] - Sr[x1-1][y2] - Sr[x2][y1-1] + Sr[x1-1][y1-1];
            ll Sc_sub = Sc[x2][y2] - Sc[x1-1][y2] - Sc[x2][y1-1] + Sc[x1-1][y1-1];

            ll W = y2 - y1 + 1;
            ll sum_val = (ll)W * (Sr_sub - (ll)x1 * S_sub) + (Sc_sub - (ll)(y1-1) * S_sub);

            s += to_string(sum_val) + ' ';
        }

        cout << 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:
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 maximize the failure probability. The bug in the code seems to be related to the calculation of the sum of the flattened submatrix, especially in the way the indices and submatrix boundaries are handled.

Let's consider the following aspects:
1. **Edge Cases for Matrix Size and Query Boundaries:**
   - Use the maximum and minimum possible values for the matrix size and query boundaries.
   - Use a matrix with a single element.
   - Use a matrix with all elements being the same.
   - Use a matrix where the submatrix spans the entire matrix.
   - Use a matrix where the submatrix is a single row or a single column.

2. **Edge Cases for Query Parameters:**
   - Use queries where \(x_1 = x_2\) and \(y_1 = y_2\).
   - Use queries where the submatrix is a single row or a single column.
   - Use queries where the submatrix is the entire matrix.
   - Use queries where the submatrix is at the boundary of the matrix.

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

```plaintext
4
1 1
1
1 1 1 1
3 3
1 2 3
4 5 6
7 8 9
1 1 3 3
1 1 1 1
1 2 3 3
2 2 2 2
2000 10
1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
1 1 1 1
2000 1 2000 10
1 1 2000 10
1 10 2000 1
1 1 2000 2000
1 1 1 1
1000 1000 2000 2000
```

This test input includes:
1. A 1x1 matrix with a single query.
2. A 3x3 matrix with queries that cover various submatrices, including the entire matrix, a single element, and submatrices at the boundaries.
3. A 2000x2000 matrix with 10 queries that test the maximum size and various submatrix configurations, including single row, single column, and the entire matrix.

This input should help expose any issues in the buggy code, especially in how it handles the indices and submatrix boundaries.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

Buggy Code:
#include "bits/stdc++.h"
#define int long long
#define uint unsigned long long
#define vi vector<int>
#define vvi vector<vi >
#define vb vector<bool>
#define vvb vector<vb >
#define fr_3(i,n) for(int i=0; i<(n); i++)
#define fb_3(i,a,n) for(int i=(a); i<=(n); i++)
#define nl cout<<"\n"
#define dbg(var) cout<<#var<<"="<<var<<" "
#define all_3(v) v.begin(),v.end()
#define srt_3(v)  sort(v.begin(),v.end())         // sort
#define mxe_3(v)  *max_element(v.begin(),v.end())     // find max element in vector
#define mne_3(v)  *min_element(v.begin(),v.end())     // find min element in vector
#define unq_3(v)  v.resize(distance(v.begin(), unique(v.begin(), v.end())));
// make sure to sort before applying unique // else only consecutive duplicates would be removed
#define bin(x,y)  bitset<y>(x)
using namespace std;
     // Hardcoded, directly change from here for functions!

int MOD=1e9+7;

void mod_add(int &a , int b) {a=((a%MOD)+(b%MOD))%MOD;}
void mo_dsub(int &a , int b) {a=((a%MOD)-(b%MOD)+MOD)%MOD;}
void mo_dmul(int &a , int b) {a=((a%MOD)*(b%MOD))%MOD;}
// ================================== take ip/op like vector,pairs directly!==================================
template<typename typC,typename typD> istream &operator>>(istream &cin,pair<typC,typD> &a) { return cin>>a.first>>a.second; }
template<typename typC> istream &operator>>(istream &cin,vector<typC> &a) { for (auto &x:a) cin>>x; return cin; }
template<typename typC,typename typD> ostream &operator<<(ostream &cout,const pair<typC,typD> &a) { return cout<<a.first<<' '<<a.second; }
template<typename typC,typename typD> ostream &operator<<(ostream &cout,const vector<pair<typC,typD>> &a) { for (auto &x:a) cout<<x<<'\n'; return cout; }
template<typename typC> ostream &operator<<(ostream &cout,const vector<typC> &a) { int n=a.size(); if (!n) return cout; cout<<a[0]; for (int i=1; i<n; i++) cout<<' '<<a[i]; return cout; }
// ===================================END Of the input module ==========================================

void funny_3(){
    // int n=1,m=0;
    // string s_3;
        int n_08o, q_08o;
        cin >> n_08o >> q_08o;

        vector<vector<int>> Munni(n_08o, vector<int>(n_08o));
        vector<vector<long long>> p_o_x(n_08o + 1, vector<long long>(n_08o + 1, 0));

        for (int i = 0; i < n_08o; ++i) {
            for (int j = 0; j < n_08o; ++j) {
                cin >> Munni[i][j];
                p_o_x[i + 1][j + 1] = Munni[i][j] + p_o_x[i + 1][j] + p_o_x[i][j + 1] - p_o_x[i][j];
            }
        }

        while (q_08o--) {
            int x10_, y10_, x20_, y20_;
            cin >> x10_ >> y10_ >> x20_ >> y20_;

            long long t_l_sum = 0;
            int index = 1;

            for (int i = x10_ - 1; i < x20_; ++i) {
                for (int j = y10_ - 1; j < y20_; ++j) {
                    t_l_sum += Munni[i][j] * index;
                    index++;
                }
            }

            cout << t_l_sum << " ";
        }
        cout << endl;
    // cin>>n;
    // vi v(n);
    // cin>>v;

}

int32_t main()
{

 ios_base::sync_with_stdio(false);
 cin.tie(NULL);

    int t_c_3= 1;
    cin >> t_c_3;
    while (t_c_3--)
    {
        funny_3();
    }
    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
3 2
1 2 3
4 5 6
7 8 9
1 1 3 3
2 2 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

Buggy Code:
t = int(input())
for _ in range(t):
    n, q = map(int, input().split())
    matrix = []
    for _ in range(n):
        row = list(map(int, input().split()))
        matrix.append(row)
    results = []
    for _ in range(q):
        x1, y1, x2, y2 = map(int, input().split())
        # 调整为 0 索引
        x1 -= 1
        y1 -= 1
        x2 -= 1
        y2 -= 1
        A = []
        for i in range(x1, x2 + 1):
            for j in range(y1, y2 + 1):
                A.append(matrix[i][j])
        total = 0
        for i, num in enumerate(A, start=1):
            total += num * i
        results.append(total)
    print(" ".join(map(str, results)))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that might cause the code to fail. Here are the steps to create such a test input:

1. **Edge Cases:**
   - Queries that involve the entire matrix.
   - Queries that involve only a single cell.
   - Queries that involve the corners of the matrix.
   - Queries that involve the edges of the matrix.

2. **Large Values:**
   - Use large values for matrix elements to test the summation logic.

3. **Multiple Queries:**
   - Ensure the test case includes multiple queries to test the loop and summation logic thoroughly.

4. **Small Matrix:**
   - Use a small matrix to keep the test input manageable and within the output length limit.

Based on these considerations, here is a small-scale test input that includes various edge cases:

```plaintext
2
4 4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
1 1 4 4
1 1 1 1
1 1 1 4
4 1 4 4
3 3
1 2 3
4 5 6
7 8 9
1 1 1 1
1 1 1 3
1 3 3 3
2 2 2 2
```

This test input includes:
- A 4x4 matrix with values from 1 to 16.
- Four queries:
  - A query for the entire matrix (1 1 4 4).
  - A query for a single cell (1 1 1 1).
  - A query for the first row (1 1 1 4).
  - A query for the last row (4 1 4 4).
- A 3x3 matrix with values from 1 to 9.
- Four queries:
  - A query for a single cell (1 1 1 1).
  - A query for the first row (1 1 1 3).
  - A query for the last row (1 3 3 3).
  - A query for the center cell (2 2 2 2).

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

Buggy Code:
t=int(input())
for i in range(t):
    (n,q)=map(int,input().split())
    mi=[]
    for j in range(n):

        rdl=input()
        mir=list(map(int,rdl.split()))
        mi.append(mir)
    qryinp=[]
    for f in range(q):
        rdll=input()
        lqry=list(map(int,rdll.split()))
        qryinp.append(lqry)

    smml=[]

    for g in range(len(qryinp)):
        x1=qryinp[g][0]
        y1=qryinp[g][1]
        x2=qryinp[g][2]
        y2=qryinp[g][3]
        lansqry=[]

        smm=0
        k=1
        for rw in range(x1-1,x2):

            for cl in range(y1-1,y2):
                # print(cl)
                lansqry.append(mi[rw][cl])
                smm+=(mi[rw][cl])*k
                k+=1
        smml.append(smm)

        # smm=0
        # for b in range(len(lansqry)):
        #     smm+=lansqry[b]*(b+1)

        # smml.append(smm)

    print(*smml)

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

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

Think step by step.

Response:
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
1 1 3 3
1 1 1 1
3 3 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

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

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(br.readLine());

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

			String[] s = br.readLine().split(" ");
			int N = Integer.parseInt(s[0]);
			int Q = Integer.parseInt(s[1]);
			long[][] p = new long[N][N];
			long[][] p2 = new long[N][N];
			long[][] p3 = new long[N][N];
			for(int j = 0; j < N; j++){
				s = br.readLine().split(" ");
				for(int k = 0; k < N; k++){
					long val = Long.parseLong(s[k]);
					p[j][k] = (j >= 1 ? p[j-1][k] : 0) +  (k >= 1 ? p[j][k-1] : 0) -
					((j >= 1 && k >= 1) ? p[j-1][k-1] : 0) + val;
					p2[j][k] = (j >= 1 ? p2[j-1][k] : 0) +  (k >= 1 ? p2[j][k-1] : 0) -
					((j >= 1 && k >= 1) ? p2[j-1][k-1] : 0) + (j * val);
					p3[j][k] = (j >= 1 ? p3[j-1][k] : 0) +  (k >= 1 ? p3[j][k-1] : 0) -
					((j >= 1 && k >= 1) ? p3[j-1][k-1] : 0) + (k * val);

				}
			}
			for(int l = 1; l <= Q; l++){
				s = br.readLine().split(" ");
				int x1 = Integer.parseInt(s[0]);
				int y1 = Integer.parseInt(s[1]);
				int x2 = Integer.parseInt(s[2]);
				int y2 = Integer.parseInt(s[3]);
				x1--; y1--; x2--; y2--;
				// First take the actual 2D prefix sum within these bounds
				long plainTotal = p[x2][y2] - (y1 >= 1 ? p[x2][y1-1] : 0) -
				(x1 >= 1 ? p[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p[x1-1][y1-1] : 0);
				long total2 = p2[x2][y2] - (y1 >= 1 ? p2[x2][y1-1] : 0) -
				(x1 >= 1 ? p2[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p2[x1-1][y1-1] : 0);
				total2 -= x1 * plainTotal;
				long total3 = p3[x2][y2] - (y1 >= 1 ? p3[x2][y1-1] : 0) -
				(x1 >= 1 ? p3[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p3[x1-1][y1-1] : 0);
				total3 -= y1 * plainTotal;
				int numCols = y2 - y1 + 1;
				System.out.print((plainTotal + total3 + numCols * total2) + " ");

			}
			System.out.println();

		}
	}
}

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

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(br.readLine());
		for(int i = 1; i <= T; i++){
			String[] s = br.readLine().split(" ");
			int N = Integer.parseInt(s[0]);
			int Q = Integer.parseInt(s[1]);
			long[][] grid = new long[N][N];
			long[][] grid2 = new long[N][N];
			long[][] grid3 = new long[N][N];
			long[][] p = new long[N][N];
			long[][] p2 = new long[N][N];
			long[][] p3 = new long[N][N];
			for(int j = 0; j < N; j++){
				s = br.readLine().split(" ");
				for(int k = 0; k < N; k++){
					grid[j][k] = Long.parseLong(s[k]);
					grid2[j][k] = j * grid[j][k];
					grid3[j][k] = k * grid[j][k];
					p[j][k] = (j >= 1 ? p[j-1][k] : 0) +  (k >= 1 ? p[j][k-1] : 0) -
					((j >= 1 && k >= 1) ? p[j-1][k-1] : 0) + grid[j][k];
					p2[j][k] = (j >= 1 ? p2[j-1][k] : 0) +  (k >= 1 ? p2[j][k-1] : 0) -
					((j >= 1 && k >= 1) ? p2[j-1][k-1] : 0) + grid2[j][k];
					p3[j][k] = (j >= 1 ? p3[j-1][k] : 0) +  (k >= 1 ? p3[j][k-1] : 0) -
					((j >= 1 && k >= 1) ? p3[j-1][k-1] : 0) + grid3[j][k];

				}
			}
			for(int l = 1; l <= Q; l++){
				s = br.readLine().split(" ");
				int x1 = Integer.parseInt(s[0]);
				int y1 = Integer.parseInt(s[1]);
				int x2 = Integer.parseInt(s[2]);
				int y2 = Integer.parseInt(s[3]);
				x1--; y1--; x2--; y2--;
				// First take the actual 2D prefix sum within these bounds
				long plainTotal = p[x2][y2] - (y1 >= 1 ? p[x2][y1-1] : 0) -
				(x1 >= 1 ? p[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p[x1-1][y1-1] : 0);
				long total2 = p2[x2][y2] - (y1 >= 1 ? p2[x2][y1-1] : 0) -
				(x1 >= 1 ? p2[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p2[x1-1][y1-1] : 0);
				total2 -= x1 * plainTotal;
				long total3 = p3[x2][y2] - (y1 >= 1 ? p3[x2][y1-1] : 0) -
				(x1 >= 1 ? p3[x1-1][y2] : 0) + ((x1 >= 1 && y1 >= 1) ? p3[x1-1][y1-1] : 0);
				total3 -= y1 * plainTotal;
				int numCols = y2 - y1 + 1;
				System.out.print((plainTotal + total3 + numCols * total2) + " ");

			}
			System.out.println();

		}
	}
} */

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that might cause issues. The primary issues in the code could arise from:

1. **Indexing and Boundary Conditions**: The code uses 0-based indexing for the matrix, but the input coordinates are 1-based. This can lead to off-by-one errors.
2. **Large Values and Overflow**: The code deals with large values and prefix sums, which can lead to overflow if not handled correctly.
3. **Complex Queries**: Queries that involve large submatrices or submatrices with specific patterns can expose issues in the prefix sum calculations.

Given these considerations, a small-scale test input that can expose these issues is:

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

This test input includes:
1. A 4x4 matrix with specific values to test the prefix sum calculations.
2. Three queries that cover different submatrices, including the entire matrix, a small submatrix, and a submatrix that spans multiple rows and columns.
3. A 3x3 matrix with a simple pattern to test basic functionality.
4. Three queries on the 3x3 matrix to ensure the code handles smaller matrices correctly.

This input is designed to test the code's ability to handle different types of queries and matrix sizes, and to ensure that the prefix sum calculations are correct.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Demon Problem
time_limit_ms: 3500
memory_limit_mb: 512
Description: Swing is opening a pancake factory! A good pancake factory must be good at flattening things, so Swing is going to test his new equipment on 2D matrices.

Swing is given an $$$n \times n$$$ matrix $$$M$$$ containing positive integers. He has $$$q$$$ queries to ask you.

For each query, he gives you four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, $$$y_2$$$ and asks you to flatten the submatrix bounded by $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ into an array $$$A$$$. Formally, $$$A = [M_{(x1,y1)}, M_{(x1,y1+1)}, \ldots, M_{(x1,y2)}, M_{(x1+1,y1)}, M_{(x1+1,y1+1)}, \ldots, M_{(x2,y2)}]$$$.

The following image depicts the flattening of a submatrix bounded by the red dotted lines. The orange arrows denote the direction that the elements of the submatrix are appended to the back of $$$A$$$, and $$$A$$$ is shown at the bottom of the image.

Afterwards, he asks you for the value of $$$\sum_{i=1}^{|A|} A_i \cdot i$$$ (sum of $$$A_i \cdot i$$$ over all $$$i$$$).

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

The first line of each test contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2000, 1 \leq q \leq 10^6$$$) — the length of $$$M$$$ and the number of queries.

The following $$$n$$$ lines contain $$$n$$$ integers each, the $$$i$$$'th of which contains $$$M_{(i,1)}, M_{(i,2)}, \ldots, M_{(i,n)}$$$ ($$$1 \leq M_{(i, j)} \leq 10^6$$$).

The following $$$q$$$ lines contain four integers $$$x_1$$$, $$$y_1$$$, $$$x_2$$$, and $$$y_2$$$ ($$$1 \leq x_1 \leq x_2 \leq n, 1 \leq y_1 \leq y_2 \leq n$$$) — the bounds of the query.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output the results of the $$$q$$$ queries on a new line.

Examples:
input:
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
output:
500 42 168
14 42 5

Note: In the second query of the first test case, $$$A = [9, 5, 5, 2]$$$. Therefore, the sum is $$$1 \cdot 9 + 2 \cdot 5 + 3 \cdot 5 + 4 \cdot 2 = 42$$$.

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

public class Main {
    static int n, q;
    static int [][] M;
    public static void main(String [] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;

        int t = Integer.parseInt(br.readLine());
        while (t --> 0) {
            st = new StringTokenizer(br.readLine());
            n = Integer.parseInt(st.nextToken());
            q = Integer.parseInt(st.nextToken());
            M = new int[n + 1][n + 1];
            for (int i = 1; i <= n; i++) {
                st = new StringTokenizer(br.readLine());
                for (int j = 1; j <= n; j++) {
                    M[i][j] = Integer.parseInt(st.nextToken());
                }
            }
            long [][] prefix1 = new long[n + 1][n + 1]; // Regular
            long [][] prefix2 = new long[n + 1][n + 1]; // Multiplied by index
            long [][] prefix3 = new long[n + 1][n + 1]; // Multiplied by row
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    prefix1[i][j] = prefix1[i - 1][j] + prefix1[i][j - 1] - prefix1[i - 1][j - 1];
                    prefix2[i][j] = prefix2[i - 1][j] + prefix2[i][j - 1] - prefix2[i - 1][j - 1];
                    prefix3[i][j] = prefix3[i - 1][j] + prefix3[i][j - 1] - prefix3[i - 1][j - 1];
                    prefix1[i][j] += M[i][j]; prefix2[i][j] += (long) M[i][j] * (i * n + j); prefix3[i][j] += (long) M[i][j] * i;
                }
            }

            while (q --> 0) {
                st = new StringTokenizer(br.readLine());
                int x1 = Integer.parseInt(st.nextToken());
                int y1 = Integer.parseInt(st.nextToken());
                int x2 = Integer.parseInt(st.nextToken());
                int y2 = Integer.parseInt(st.nextToken());

                long res = sum(prefix2, x1, y1, x2, y2);
                res -= sum(prefix3, x1, y1, x2, y2) * (n - (y2 - y1 + 1));
                res += sum(prefix1, x1, y1, x2, y2) * (x1 * (n - (y2 - y1 + 1)) - (x1 * n + y1 - 1));
                System.out.print(res + " ");
            }
            System.out.println();
        }

        br.close();
    }

    public static long sum(long [][] prefix, int l1, int r1, int l2, int r2) {
        return prefix[l2][r2] - prefix[l1 - 1][r2] - prefix[l2][r1 - 1] + prefix[l1 - 1][r1 - 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
2
4 3
1 5 2 4
4 9 5 3
4 5 2 3
1 5 5 2
1 1 4 4
2 2 3 3
1 2 4 3
3 3
1 2 3
4 5 6
7 8 9
1 1 1 3
1 3 3 3
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Phone Desktop
time_limit_ms: 1000
memory_limit_mb: 256
Description: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \times 3$, i.e., five rows and three columns.

There are $x$ applications with an icon size of $1 \times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.

Rosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.

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

The first and only line of each test case contains two integers $x$ and $y$ ($0 \leq x, y \leq 99$) — the number of applications with a $1 \times 1$ icon and the number of applications with a $2 \times 2$ icon, respectively.

Output: For each test case, output the minimal number of required screens on a separate line.

Examples:
input:
11
1 1
7 2
12 4
0 3
1 0
8 1
0 0
2 0
15 0
8 2
0 9
output:
1
1
2
2
1
1
0
1
1
2
5

Note: The solution for the first test case can look as follows:

![](CDN_BASE_URL/c1f8e47d264264fb303b101863bef14b) Blue squares represent empty spaces for icons, green squares represent $1 \times 1$ icons, red squares represent $2 \times 2$ icons

The solution for the third test case can look as follows:

![](CDN_BASE_URL/05b392e954bd162bfc00b3f026bcba0d)

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

#define X 15
#define Y 2

int main() {
	int t; cin >> t;
	while (t--) {
		int x, y; cin >> x >> y;
		int cnt = 0;
		cnt += y / Y;
		if (y % 2) cnt++;
		int num = cnt * X - y * 4;
		if (x <= num) cout << cnt << endl;
		else {
			x -= num;
			if (x <= X) cout << cnt + 1 << endl;
			else cout << x / X + 1 << endl;
		}
	}
	return 0;
}

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

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

Think step by step.

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

1. **Maximum Number of Large Icons (2x2)**: The code might handle a large number of 2x2 icons incorrectly.
2. **Mix of Large and Small Icons**: The code might not correctly handle a combination of 1x1 and 2x2 icons, especially when the number of 1x1 icons is just enough to fill the remaining space after placing 2x2 icons.
3. **No Icons**: The code should handle the case where there are no icons (both x and y are 0).
4. **Single Icon**: The code should handle the case where there is only one icon, either 1x1 or 2x2.
5. **Exact Fit**: The code should handle the case where the icons fit exactly into the screens without any leftover space.

Let's construct a small-scale test input that covers these scenarios:

```plaintext
5
0 9
15 0
7 2
1 1
0 0
```

### Explanation:
1. **0 9**: This case tests the maximum number of 2x2 icons. The code should correctly handle the placement of 9 2x2 icons, which requires 5 screens.
2. **15 0**: This case tests a large number of 1x1 icons. The code should correctly handle the placement of 15 1x1 icons, which requires 1 screen.
3. **7 2**: This case tests a mix of 1x1 and 2x2 icons. The code should correctly handle the placement of 7 1x1 icons and 2 2x2 icons, which requires 1 screen.
4. **1 1**: This case tests a single 1x1 icon and a single 2x2 icon. The code should correctly handle the placement of 1 1x1 icon and 1 2x2 icon, which requires 1 screen.
5. **0 0**: This case tests the scenario where there are no icons. The code should correctly handle this and output 0 screens.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Phone Desktop
time_limit_ms: 1000
memory_limit_mb: 256
Description: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \times 3$, i.e., five rows and three columns.

There are $x$ applications with an icon size of $1 \times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.

Rosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.

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

The first and only line of each test case contains two integers $x$ and $y$ ($0 \leq x, y \leq 99$) — the number of applications with a $1 \times 1$ icon and the number of applications with a $2 \times 2$ icon, respectively.

Output: For each test case, output the minimal number of required screens on a separate line.

Examples:
input:
11
1 1
7 2
12 4
0 3
1 0
8 1
0 0
2 0
15 0
8 2
0 9
output:
1
1
2
2
1
1
0
1
1
2
5

Note: The solution for the first test case can look as follows:

![](CDN_BASE_URL/c1f8e47d264264fb303b101863bef14b) Blue squares represent empty spaces for icons, green squares represent $1 \times 1$ icons, red squares represent $2 \times 2$ icons

The solution for the third test case can look as follows:

![](CDN_BASE_URL/05b392e954bd162bfc00b3f026bcba0d)

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
#define c1 cout<< -1 <<endl
#define PI 3.141592653589793238462643383279
const ll mod= 1e9 + 7;

ll BinExp(ll a,ll b,ll mod){ll ans=1;while(b>0){if(b&1){ans=(ans*1LL*a)%mod;}a=(a*1LL*a)%mod;b>>=1;}return ans;}
ll m_mod(ll x, ll y) { return ((x % mod) * (y % mod)) % mod;}
ll m_divmod(ll x,ll y,ll m){return (x%mod*BinExp(y,m-2,m))%m;}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin>>t;
    while(t--){
      int x,y;
      cin>>x>>y;
      int ct,yt;
      if(y==0){
        if(x==0){
          cout<<0<<endl;
        }
        else{
          ct = (int)(x/15);
          if(x%15 != 0){
            ct++;
          }
          cout<<ct<<endl;
        }
      }
      else if(x==0){
        yt = (int)(y/2);
        if(y%2 != 0){
          yt++;
        }
        cout<<yt<<endl;
      }
      else{
        ct = (int)(x/7);
        yt = (int)(y/2);
        if(x%7 != 0){
          ct++;
        }
        if(y%2 != 0){
          yt++;
        }
        if(yt>=ct){
          cout<<yt<<endl;
        }
        else{
          y = y%2;
          x = x%7;
          if(y==1){
            if(x<=4){
              cout<<yt<<endl;
            }
            else{
              cout<<ct<<endl;
            }
          }
          else{
            cout<<ct<<endl;
          }
        }
      }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1
14 2
15 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Phone Desktop
time_limit_ms: 1000
memory_limit_mb: 256
Description: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \times 3$, i.e., five rows and three columns.

There are $x$ applications with an icon size of $1 \times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.

Rosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.

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

The first and only line of each test case contains two integers $x$ and $y$ ($0 \leq x, y \leq 99$) — the number of applications with a $1 \times 1$ icon and the number of applications with a $2 \times 2$ icon, respectively.

Output: For each test case, output the minimal number of required screens on a separate line.

Examples:
input:
11
1 1
7 2
12 4
0 3
1 0
8 1
0 0
2 0
15 0
8 2
0 9
output:
1
1
2
2
1
1
0
1
1
2
5

Note: The solution for the first test case can look as follows:

![](CDN_BASE_URL/c1f8e47d264264fb303b101863bef14b) Blue squares represent empty spaces for icons, green squares represent $1 \times 1$ icons, red squares represent $2 \times 2$ icons

The solution for the third test case can look as follows:

![](CDN_BASE_URL/05b392e954bd162bfc00b3f026bcba0d)

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

// Macros
#define fast_io ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define ll long long
#define ull unsigned long long
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int, int>
#define pll pair<long long, long long>
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define sz(x) (int)(x).size()
#define rep(i,a,b) for(int i = a; i < b; ++i)
#define rrep(i,a,b) for(int i = a; i >= b; --i)

// Constants
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ll LINF = 1e18;
const double EPS = 1e-9;

int ceil_div(int a, int b) {
    return (a + b - 1) / b;
}

void solve() {
    int a, b;
    cin >> a >> b;
    int count = 0;
    if(b==0){
      count=ceil_div(a,15);
    }
    if(b%2==1){
      count=b/2+1;
      int x=9+7*(count);
      if(a>x){
        count+=ceil_div(a-x,15);
      }
    }
    if(b%2==0){
      count=b/2;
      int x=7*(count);

      if(a>x){
        count+=ceil_div(a-x,15);
      }
    }

    cout << count << endl;
}

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:
```plaintext
1
14 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Phone Desktop
time_limit_ms: 1000
memory_limit_mb: 256
Description: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \times 3$, i.e., five rows and three columns.

There are $x$ applications with an icon size of $1 \times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.

Rosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.

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

The first and only line of each test case contains two integers $x$ and $y$ ($0 \leq x, y \leq 99$) — the number of applications with a $1 \times 1$ icon and the number of applications with a $2 \times 2$ icon, respectively.

Output: For each test case, output the minimal number of required screens on a separate line.

Examples:
input:
11
1 1
7 2
12 4
0 3
1 0
8 1
0 0
2 0
15 0
8 2
0 9
output:
1
1
2
2
1
1
0
1
1
2
5

Note: The solution for the first test case can look as follows:

![](CDN_BASE_URL/c1f8e47d264264fb303b101863bef14b) Blue squares represent empty spaces for icons, green squares represent $1 \times 1$ icons, red squares represent $2 \times 2$ icons

The solution for the third test case can look as follows:

![](CDN_BASE_URL/05b392e954bd162bfc00b3f026bcba0d)

Buggy Code:
t = int(input())
if (t>=1) and (t<=10000) :
    A = []
    for i in range(t) :
        a = []
        x,y = map(int, input().split())
        a.append(x)
        a.append(y)
        A.append(a)

    for i in range(len(A)) :
        if (A[i][0]>=0) and (A[i][0]<=99) and (A[i][1]>=0) and (A[i][1]<=99) :
            count = A[i][0] + A[i][1]*4
            if A[i][1]%2==0 :
                if A[i][1]==0 :
                    if A[i][0]%15==0 :
                        screens = A[i][0]/15
                    else :
                        screens = A[i][0]//15 + 1
                else :
                    screens = A[i][1]/2
            else :
                screens = A[i][1]//2 + 1
            if (A[i][0]>((screens*15)-(A[i][1]*4))) :
                screens += (A[i][0] - (screens*7))//15 + 1
                print(int(screens))
            else :
                print(int(screens))
        else :
            print("Error")
else :
    print("Error")

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that the code might not handle correctly. The code seems to have issues with the logic for calculating the number of screens, especially when dealing with the placement of both 1x1 and 2x2 icons. Here are some key scenarios to consider:

1. **All 2x2 Icons**: This scenario tests the code's ability to handle only 2x2 icons.
2. **All 1x1 Icons**: This scenario tests the code's ability to handle only 1x1 icons.
3. **Mixed Icons with More 2x2 Icons**: This scenario tests the code's ability to handle a mix of 1x1 and 2x2 icons, where the number of 2x2 icons is significant.
4. **Mixed Icons with More 1x1 Icons**: This scenario tests the code's ability to handle a mix of 1x1 and 2x2 icons, where the number of 1x1 icons is significant.
5. **Edge Cases with Small Numbers**: This scenario tests the code's ability to handle small numbers of icons, which can sometimes expose logical errors.

Given these considerations, let's create a small-scale test input that includes these scenarios:

```plaintext
5
0 3
15 0
12 4
7 2
1 1
```

### Explanation:
1. **0 3**: All 2x2 icons. The correct output should be 2 screens.
2. **15 0**: All 1x1 icons. The correct output should be 1 screen.
3. **12 4**: Mixed icons with more 2x2 icons. The correct output should be 2 screens.
4. **7 2**: Mixed icons with a balance of 1x1 and 2x2 icons. The correct output should be 1 screen.
5. **1 1**: Mixed icons with small numbers. The correct output should be 1 screen.

This test input should help expose any issues in the logic of the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Phone Desktop
time_limit_ms: 1000
memory_limit_mb: 256
Description: Little Rosie has a phone with a desktop (or launcher, as it is also called). The desktop can consist of several screens. Each screen is represented as a grid of size $5 \times 3$, i.e., five rows and three columns.

There are $x$ applications with an icon size of $1 \times 1$ cells; such an icon occupies only one cell of the screen. There are also $y$ applications with an icon size of $2 \times 2$ cells; such an icon occupies a square of $4$ cells on the screen. Each cell of each screen can be occupied by no more than one icon.

Rosie wants to place the application icons on the minimum number of screens. Help her find the minimum number of screens needed.

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

The first and only line of each test case contains two integers $x$ and $y$ ($0 \leq x, y \leq 99$) — the number of applications with a $1 \times 1$ icon and the number of applications with a $2 \times 2$ icon, respectively.

Output: For each test case, output the minimal number of required screens on a separate line.

Examples:
input:
11
1 1
7 2
12 4
0 3
1 0
8 1
0 0
2 0
15 0
8 2
0 9
output:
1
1
2
2
1
1
0
1
1
2
5

Note: The solution for the first test case can look as follows:

![](CDN_BASE_URL/c1f8e47d264264fb303b101863bef14b) Blue squares represent empty spaces for icons, green squares represent $1 \times 1$ icons, red squares represent $2 \times 2$ icons

The solution for the third test case can look as follows:

![](CDN_BASE_URL/05b392e954bd162bfc00b3f026bcba0d)

Buggy Code:
def box(x,y):
    if x <= 0 and y <= 0:
        return 0
    if x > 0 and y <= 0:
        return 1 + box(x-15,y)
    elif x <= 0 and y >0:
        return 1 + box(x,y-2)
    else:
        if y <= 2:
            if x + 4*y <= 15: return 1
            else: return 1 + box(x-7,0)
        elif y > 2:
            return 1 + box(x-7,y-2)
n = int(input())
for i in range(n):
     k = input()
     s = []
     for j in k.split():
         s.append(int(j))
     print(box(s[0],s[1]))

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

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

Think step by step.

Response:
```plaintext
1
14 3
```