**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fi first
#define se second
#define PII pair<int,int>
#define endl '\n'
#define lowbit(x) x & (-x)

const int N = 1e5+5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f, Mod = 998244353;
int n, m;

void solve()
{
	cin>>n>>m;
	ll x, y;
	cin>>x>>y;
	int x1 = n - x;
	ll res = (ll)n * (m - x + 1) - y - x1 + 1ll * x1 * m;
	cout<<res<<endl;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int t = 1;
	cin>>t;
	while(t--)
		solve();

	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases or scenarios that might cause the code to fail. Let's analyze the code and the problem statement to identify potential issues.

### Analysis of the Buggy Code
1. **Input Reading:**
   - The code reads `n`, `m`, `x`, and `y` correctly.
2. **Calculation:**
   - The code calculates `x1` as `n - x`.
   - The result is calculated using the formula:
     \[
     \text{res} = (n \cdot (m - x + 1)) - y - x1 + (x1 \cdot m)
     \]
   - This formula seems incorrect, as it does not correctly account for the Manhattan distances of the movements.

### Potential Issues
1. **Incorrect Calculation of `x1`:**
   - `x1` should be `n - x + 1` instead of `n - x` to correctly account for the number of rows below the row where the person left.
2. **Incorrect Formula:**
   - The formula for `res` is not correctly reflecting the sum of the Manhattan distances.

### Test Input to Expose the Bug
To expose the bug, we can use a test case where the person leaves from a position that is not at the edge of the grid. This will make the incorrect formula more apparent.

Let's use the following test input:
```plaintext
1
3 4 2 3
```

### Explanation
- **Grid Size:** 3 rows and 4 columns.
- **Person Leaving:** Position (2, 3), which corresponds to the person numbered 7 in row-major order.
- **Expected Movement:**
  - Person 8 moves to position (2, 2) (Manhattan distance = 1)
  - Person 9 moves to position (2, 3) (Manhattan distance = 1)
  - Person 10 moves to position (2, 4) (Manhattan distance = 1)
  - Person 11 moves to position (3, 1) (Manhattan distance = 2)
  - Person 12 moves to position (3, 2) (Manhattan distance = 2)
- **Expected Sum of Manhattan Distances:**
  \[
  1 + 1 + 1 + 2 + 2 = 7
  \]

### Expected Output
The correct output for the test input should be:
```plaintext
7
```

### Conclusion
The test input:
```plaintext
1
3 4 2 3
```
will expose the bug in the given code, as the incorrect formula will produce a different result from the expected output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
for _ in range(int(input())):
    n, m, r, c = map(int, input().split())
    person = (r-1)*m + c
    row_num = person // m + 1
    tot = n * m
    x = tot - (n - row_num) - person

    # print(person, diag, tot, x)

    sum_total = ((n - row_num) * m) + (1 * x)

    print(sum_total)

**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 given code, we need to focus on edge cases where the code might fail. Let's analyze the code and identify potential issues:

1. **Row and Column Calculation**: The code calculates the row number using `row_num = person // m + 1`. This might be incorrect if the person number is at the boundary of a row.
2. **Total Movements**: The calculation of `x` and `sum_total` might not correctly account for the number of people moving and their distances.

Given these points, let's create a test case that highlights these issues:

- **Test Case**: Consider a scenario where the person leaving is at the last position of a row. This will test the row number calculation and the movement distances.

### Test Case
```plaintext
1
3 4 3 4
```

### Explanation
- **Input**: `3 4 3 4`
  - `n = 3` (3 rows)
  - `m = 4` (4 columns)
  - `r = 3` (3rd row)
  - `c = 4` (4th column)

- **Person Number**: The person at position (3, 4) is numbered `(3-1)*4 + 4 = 12`.

- **Expected Behavior**:
  - The person numbered 12 leaves.
  - The remaining people are numbered 13, 14, 15, and 16.
  - They will move to the positions of 12, 13, 14, and 15 respectively.
  - The distances are:
    - Person 13 moves from (3, 1) to (3, 4): `|3-3| + |1-4| = 3`
    - Person 14 moves from (3, 2) to (3, 1): `|3-3| + |2-1| = 1`
    - Person 15 moves from (3, 3) to (3, 2): `|3-3| + |3-2| = 1`
    - Person 16 moves from (3, 4) to (3, 3): `|3-3| + |4-3| = 1`

- **Expected Output**: `3 + 1 + 1 + 1 = 6`

This test case should expose any issues in the row number calculation and the total movement distance calculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
import math

def Sol():

    n, m, r, c = map(int, input().strip().split())
    pos = (r - 1) * m + c
    ans = (n - r) * m + (m - c) * n + (pos % m) - r
    print(max(0, ans))
    return

for _ in range(int(input())):
    Sol()

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose the bug in the given buggy code. Let's break down why this specific input is challenging:

1. **Small-Scale Input**: The input is small enough to be easily understood and manually verified, but it is still capable of exposing the bug.

2. **Position to Leave**: The person leaving is at position (1, 2), which means the person numbered 2 leaves. This leaves a gap in the first row, second column.

3. **Movement of People**:
   - The person numbered 3 (initially at (2, 1)) will move to (1, 2).
   - The person numbered 4 (initially at (2, 2)) will move to (2, 1).

4. **Manhattan Distances**:
   - Person 3 moves from (2, 1) to (1, 2): Manhattan distance = |2-1| + |1-2| = 1 + 1 = 2.
   - Person 4 moves from (2, 2) to (2, 1): Manhattan distance = |2-2| + |2-1| = 0 + 1 = 1.

5. **Expected Output**: The sum of the Manhattan distances is 2 + 1 = 3.

### Bug Exposure:
The buggy code calculates the answer as:
```python
ans = (n - r) * m + (m - c) * n + (pos % m) - r
```
For the input (2, 2, 1, 2), the calculation is:
- \( pos = (1 - 1) * 2 + 2 = 2 \)
- \( ans = (2 - 1) * 2 + (2 - 2) * 2 + (2 % 2) - 1 = 2 + 0 + 0 - 1 = 1 \)

The buggy code outputs 1, which is incorrect. The correct output should be 3.

This test case effectively demonstrates the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

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

    cal=(r-1)*m+c
    val=n*m-cal

    we=val//n
    ans=we*n-we*1+val
    print(ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code calculates the sum of Manhattan distances incorrectly, particularly when the person leaving is in the last row or when the grid is very large.

Let's analyze the code:
1. The code calculates the position of the person who leaves using `cal = (r-1)*m + c`.
2. It then calculates the number of people who need to move using `val = n*m - cal`.
3. It computes the number of full rows that need to be moved using `we = val // n`.
4. Finally, it calculates the total distance using `ans = we*n - we*1 + val`.

The main issues with this code are:
- It does not correctly account for the movement of people in the last row.
- It does not handle the case where the person leaving is in the last row or column correctly.

To expose these issues, we can use the following test cases:

1. **Person leaving is in the last row and last column:**
   - This will test the edge case where the person leaving is at the bottom-right corner of the grid.
   - Example: `n = 3, m = 3, r = 3, c = 3`

2. **Person leaving is in the last row but not the last column:**
   - This will test the edge case where the person leaving is in the last row but not at the end.
   - Example: `n = 3, m = 3, r = 3, c = 2`

3. **Person leaving is in the last column but not the last row:**
   - This will test the edge case where the person leaving is in the last column but not at the bottom.
   - Example: `n = 3, m = 3, r = 2, c = 3`

4. **Person leaving is in the first row and first column:**
   - This will test the edge case where the person leaving is at the top-left corner of the grid.
   - Example: `n = 3, m = 3, r = 1, c = 1`

5. **Person leaving is in the middle of the grid:**
   - This will test a general case where the person leaving is not at the edge of the grid.
   - Example: `n = 3, m = 3, r = 2, c = 2`

Here is the test input:

```plaintext
5
3 3 3 3
3 3 3 2
3 3 2 3
3 3 1 1
3 3 2 2
```

This test input covers a variety of edge cases and a general case, which should help in exposing the bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
import java.util.*;
public class Solution{
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t>0){
            long n=sc.nextInt();
            long m=sc.nextInt();
            long r=sc.nextInt();
            long c=sc.nextInt();

            long l=n*m;
            long i=(r-1)*m+c;
            long dist=0;

            long total=l/m;

            long done=(i/m)+1;

            long rem=total-done;

            long remL=l-i;

            long ans=(remL-rem)+(rem*m);

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {

        List<List<Integer>> data = readInput();
        System.out.println("6\n" +
                "1\n" +
                "0\n" +
                "1999998000000");
    }

    // read input
    static List<List<Integer>> readInput() {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();

        List<List<Integer>> input = new ArrayList<>();
        for (int i = 0; i < n; i++){
            List<Integer> testcase = new ArrayList<>();
            for (int y = 0; y < 4; y++) {
                int element = scanner.nextInt();
                testcase.add(element);
            }
            input.add(testcase);
        }
        return input;
    }
}

**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 create a test input that highlights the issue. The buggy code does not actually compute the sum of the Manhattan distances; it simply prints a hardcoded output. This means any input that deviates from the hardcoded output will reveal the bug.

### Step-by-Step Analysis:
1. **Identify the Hardcoded Output:**
   - The hardcoded output is:
     ```
     6
     1
     0
     1999998000000
     ```
   - This output corresponds to the following test cases:
     - `2 3 1 2` -> 6
     - `2 2 2 1` -> 1
     - `1 1 1 1` -> 0
     - `1000000 1000000 1 1` -> 1999998000000

2. **Create a New Test Case:**
   - We need a test case that is not one of the above to ensure the code fails.
   - A good candidate is a small-scale test case that is simple to verify but different from the hardcoded ones.

3. **Choose a Test Case:**
   - Let's choose a simple test case with a small grid and a different position:
     - `n = 2`, `m = 2`, `r = 1`, `c = 1`

4. **Verify the Correct Output:**
   - For the test case `2 2 1 1`:
     - The person at position (1, 1) leaves, which is person number 1.
     - The remaining people are numbered 2 and 3.
     - Person 2 moves from (1, 2) to (1, 1), distance = 1.
     - Person 3 moves from (2, 1) to (1, 2), distance = 1.
     - Person 4 moves from (2, 2) to (2, 1), distance = 1.
     - Total distance = 1 + 1 + 1 = 3.

### Test Input:
```plaintext
1
2 2 1 1
```

This test input should cause the buggy code to fail because it does not compute the correct sum of Manhattan distances, and the output will not match the expected result of 3.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    long long t;
    cin>>t;
    while(t--){
        long long n,m,r,c;
        cin>>n>>m>>r>>c;
        long long a;
        a=(r-1)*m+c;
        long long f=(a+n-1)/n;
        long long t=a%m;
        long long ans;
        if(t==0){
        ans=(n-f)*m+(n-f)*(m-1);
        }
        else{
              ans=(m-t)+(n-f)*m+(n-f)*(m-1);
        }
        cout<<ans<<endl;

    }
    return 0;
}

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

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

Think step by step.

Response:
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. The code calculates the sum of the Manhattan distances when a person leaves a specific position in a grid. Let's break down the potential issues and edge cases:

1. **Single Row or Column**: When the grid has only one row or one column, the movement logic can be different.
2. **Large Grid**: Large values of \( n \) and \( m \) can cause overflow or incorrect calculations.
3. **Boundary Conditions**: When the person leaves from the first or last row or column.
4. **General Case**: A general case that might not be handled correctly by the code.

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

```plaintext
6
1 3 1 2
2 1 2 1
1000000 1000000 1000000 1000000
3 3 1 1
3 3 3 3
1 1 1 1
```

Explanation:
1. **Single Row (1 3 1 2)**: The person in the second position of a single row grid leaves. This tests the boundary condition for a single row.
2. **Single Column (2 1 2 1)**: The person in the second position of a single column grid leaves. This tests the boundary condition for a single column.
3. **Large Grid (1000000 1000000 1000000 1000000)**: The person in the last position of a large grid leaves. This tests the handling of large values.
4. **Top-Left Corner (3 3 1 1)**: The person in the top-left corner of a 3x3 grid leaves. This tests the movement logic from the top-left corner.
5. **Bottom-Right Corner (3 3 3 3)**: The person in the bottom-right corner of a 3x3 grid leaves. This tests the movement logic from the bottom-right corner.
6. **Single Person (1 1 1 1)**: The only person in a 1x1 grid leaves. This tests the edge case of a single person.

These test cases should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
#include <iostream>

using namespace std;

int main()
{
    int n;
    cin>>n;
    int i = 0;
    long int res[n];
    while(n>0){
        long long int arr[4];
        for(int i = 0; i<4; i++){
            cin>>arr[i];
        }
        long long int total = arr[0]*arr[1];
        long long int pre = ((arr[2]-1)*(arr[0])+ (arr[3]-1));
        //cout<<pre<<endl;
        long long int remain = total-(pre+1);
        //cout<<remain<<endl;
        long long int result;
        if(remain<arr[1]){
            result = remain;
        }
        else{
            result = remain+((arr[1]-arr[3])*(arr[1]-1));
        }
        cout<<result<<endl;
        i++;
        n--;
    }
    /*for(int j = 0; j<n; j++){
        cout<<res[j]<<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 expose the bug in the provided code, let's consider a few key points:

1. **Edge Cases:**
   - When the grid is very small (e.g., 1x1, 1x2, 2x1).
   - When the person leaving is in the last row or last column.
   - When the person leaving is in the first row or first column.

2. **Large Input:**
   - When the grid is very large (e.g., 10^6 x 10^6).
   - When the person leaving is near the middle of the grid.

3. **Boundary Conditions:**
   - When the grid has only one row or one column.
   - When the person leaving is at the edge of the grid.

Let's create a test case that includes a combination of these scenarios to maximize the chance of exposing the bug.

### Test Case:
- **Small Grid:**
  - 1x1 grid: Person leaves from (1,1).
  - 2x2 grid: Person leaves from (1,1), (1,2), (2,1), (2,2).

- **Large Grid:**
  - 10^6 x 10^6 grid: Person leaves from (1,1), (1,10^6), (10^6,1), (10^6,10^6).

- **Mixed Grid:**
  - 3x3 grid: Person leaves from (1,1), (1,3), (3,1), (3,3).

### Test Input:
```plaintext
12
1 1 1 1
2 2 1 1
2 2 1 2
2 2 2 1
2 2 2 2
1000000 1000000 1 1
1000000 1000000 1 1000000
1000000 1000000 1000000 1
1000000 1000000 1000000 1000000
3 3 1 1
3 3 1 3
3 3 3 1
3 3 3 3
```

This test input includes a variety of edge cases and large inputs to ensure the code is thoroughly tested for potential bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
import sys
input = lambda: sys.stdin.readline().rstrip()
ii = lambda: int(input())
mii = lambda: map(int, input().split())

def solve():
    m, n, r, c = mii()
    r -= 1; c -= 1
    return (m-1-r) * (n-1) + m*n - (r*m+c) - 1

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

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

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

Think step by step.

Response:
```plaintext
4
2 3 2 3
1 2 1 2
3 3 3 3
1000000 2 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
no_tests = int(input())

while (no_tests > 0):
    rows, columns, removed_row, removed_column = map(int, input().split(" "))
    total1 = total2 = 0
    # to_first =
    # for i in range((removed_row - 1) * columns + removed_column + 1, rows * columns + 1):
    #     current_column = (i - 1) % columns + 1
    #     current_row = i // columns

    #     next_row = current_row
    #     next_column = current_column - 1
    #     if next_column <= 0:
    #         # Roll over
    #         next_column = columns
    #         next_row = current_row - 1

    #     total1 += abs(next_row - current_row) + abs(current_column - next_column)

    to_start = removed_column + columns - 1
    lr = rows - removed_row
    if removed_column == 1:
        to_start = 0
        lr += 1

    # if removed_row == 1 and not to_start:
    #     lr = rows

    lc = columns
    total2 += abs((lr) * (lc - 1)) + max(0, (lr - 1) * lc) + to_start

    # - ij - - -
    # r = 4
    # c = 5
    # - - p - -  (c - 1) + 1 = c +  c0 -1
    # ij - - - - (c0)  (r)dc = c - 1
    # ij - - - - (1)  (r-1)dl, dl = c0 - 1 + 1
    # - - - - -
    # - - - - -

    # * *
    # * +  # lr = 0

    # total distance = r(c0 - 1) + (r-1) dl
    print(total2)

    no_tests -= 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
4
1 1 1 1
2 2 1 2
3 3 2 1
1000000 1000000 1000000 1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
import java.util.Scanner;
public class Main{
    public static void main(String[] args){
        Scanner sc= new Scanner(System.in);
        int t=sc.nextInt();
        for(int k=0;k<t;k++){
            int n=sc.nextInt();
            int m=sc.nextInt();
            int r=sc.nextInt();
            int c=sc.nextInt();
            long cn=0;
            if(r<n){
                cn=(m-c)+(n-r);
                cn=cn*m;
            }
            else if(n-r==0){
                cn=m-c;
            }
            else{
                cn=(m-c)+(n-r);
                cn=cn*m;
            }
            System.out.println(cn);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
4
2 3 2 1
2 3 2 2
2 3 2 3
1000000 1000000 1000000 1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
import java.util.Scanner;

public class ManhattanDistanceSum {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int totalHuman=sc.nextInt();
        for(int i=0;i<totalHuman;i++){
            long rows=sc.nextLong();
            long col=sc.nextLong();
            long r=sc.nextLong();
            long c=sc.nextLong();
            long num=(rows - r) * (col- 1) + col * rows -  (r - 1) * rows - c;
            System.out.println(num);
        }
    }
}
//一共有n行, 每行m个人(列数)
//按行从1给每个人顺次进行编号
//(2,3)->(r,c) m=6
//编号为(2-1)*3+3=6
//最后输出曼哈顿距离的总和: abs(r0-r1)+abs(c0-r1)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code seems to have issues with the calculation of the sum of Manhattan distances. Let's break down the potential issues and generate a test case accordingly.

### Analysis of the Buggy Code
The buggy code calculates the sum of the Manhattan distances using the following formula:
```java
long num = (rows - r) * (col - 1) + col * rows - (r - 1) * rows - c;
```
This formula does not correctly compute the sum of the Manhattan distances. The correct approach should involve calculating the distance for each person who moves and summing these distances.

### Edge Cases to Consider
1. **Single Row or Column**: When the grid has only one row or one column, the movement pattern is simplified.
2. **Person at the Edge**: When the person who leaves is at the edge of the grid, the movement pattern is also simpler.
3. **Person in the Middle**: When the person who leaves is in the middle of the grid, the movement pattern is more complex and can expose issues in the calculation.

### Challenging Test Input
Let's generate a test input that includes a combination of these edge cases to maximize the failure probability.

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

### Explanation
1. **1 5 1 3**: A single row with 5 columns, and the person at position (1, 3) leaves. This tests the case where the grid has only one row.
2. **5 1 3 1**: A single column with 5 rows, and the person at position (3, 1) leaves. This tests the case where the grid has only one column.
3. **3 3 2 2**: A 3x3 grid with the person at position (2, 2) leaving. This tests the case where the person is in the middle of the grid.
4. **10 10 5 5**: A 10x10 grid with the person at position (5, 5) leaving. This tests a larger grid and a more complex movement pattern.

This test input should help expose any issues in the buggy code's calculation of the sum of Manhattan distances.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
# include <iostream>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        int n,m,r,c;
        cin>>n>>m>>r>>c;
        unsigned long long mid_calc = (r-1)*m;
        unsigned long long k = mid_calc+c;
        unsigned long long row = (k-1)/m;
        unsigned long long col = (k-1)%m;
        unsigned long long ans = m-col-1;
        ans += (2*m-1)*(n-(k-1)/m-1);
        cout<<ans<<endl;
    }
return 0;}

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

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

Think step by step.

Response:
```plaintext
3
2 3 2 3
1 1000000 1 1000000
1000000 1 1000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

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

long long getManhattanSum(long long n, long long m, long long r, long long c) {

    r--; c--;
    long long pos = r * m + c;
    long long remaining = (n * m) - (pos + 1);
    if (remaining <= 0) return 0;
    if (r == 0 && c == 0 && n == 1000000 && m == 1000000) {
        return 1999998000000LL;
    }

    long long sum = 0;
    for (long long curr = pos + 1; curr < n * m; curr++) {
        long long curr_r = curr / m;
        long long curr_c = curr % m;
        long long target = curr - 1;
        long long target_r = target / m;
        long long target_c = target % m;
        sum += abs(curr_r - target_r) + abs(curr_c - target_c);
    }

    return sum;
}

void solve() {
    long long n, m, r, c;
    cin >> n >> m >> r >> c;
    cout << getManhattanSum(n, m, r, c) << "\n";
}

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

    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
1000000 2 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
t=int(input())
ans=[]
def sum_finder(n,m,r,c):
    sum=0
    sum+=m-c
    for k in range (r+1,n+1):
        sum+=m+m-1
    ans.append(sum)
for i in range(t):
    nums=list(map(int,input().split()))
    n=nums[0]
    m=nums[1]
    r=nums[2]
    c=nums[3]
    sum_finder(n,m,r,c)
for g in range(len(ans)):
    print(ans[g])

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

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

Think step by step.

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

1. **Edge Cases**:
   - Single row or single column scenarios.
   - The person leaving is the last person in the row.
   - The person leaving is the last person in the grid.

2. **General Cases**:
   - The person leaving is in the middle of the grid.
   - The person leaving is in the first row but not the first column.
   - The person leaving is in the first column but not the first row.

3. **Large Input**:
   - Large values of \( n \) and \( m \) to test the performance and correctness of the algorithm.

Given the constraints and the need for a small-scale test input, we will focus on the edge cases and a few general cases to ensure the code is thoroughly tested.

### Test Input
```plaintext
6
1 1 1 1
1 2 1 2
2 1 2 1
2 3 1 1
2 3 1 2
2 3 1 3
```

### Explanation
- **Test Case 1**: \( n=1 \), \( m=1 \), \( r=1 \), \( c=1 \)
  - Only one person, so no movement. Expected output: 0.
- **Test Case 2**: \( n=1 \), \( m=2 \), \( r=1 \), \( c=2 \)
  - Person 2 leaves, and person 1 does not move. Expected output: 0.
- **Test Case 3**: \( n=2 \), \( m=1 \), \( r=2 \), \( c=1 \)
  - Person 2 leaves, and person 1 does not move. Expected output: 0.
- **Test Case 4**: \( n=2 \), \( m=3 \), \( r=1 \), \( c=1 \)
  - Person 1 leaves, and the movements are: 2 -> 1, 3 -> 2, 4 -> 3, 5 -> 4, 6 -> 5. Expected output: 10.
- **Test Case 5**: \( n=2 \), \( m=3 \), \( r=1 \), \( c=2 \)
  - Person 2 leaves, and the movements are: 3 -> 2, 4 -> 3, 5 -> 4, 6 -> 5. Expected output: 6.
- **Test Case 6**: \( n=2 \), \( m=3 \), \( r=1 \), \( c=3 \)
  - Person 3 leaves, and the movements are: 4 -> 3, 5 -> 4, 6 -> 5. Expected output: 3.

This test input covers a variety of scenarios, including edge cases and general cases, to ensure the code is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
t = int(input())

for i in range(t):

    n, m, r, c = map(int, input().split())

    ans = []
    for j in range(c, m):
        ans.append(1)

    if r != n:
        ans.append(m)

    sm = sum(ans)

    for j in range(r, n):
        ans.append(m - 1)
        if j != n - 1:
            ans.append(m)
    print(sum(ans))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
import java.util.*;
public class Solver {
    public static void main(String[] args) {
        Scanner SC = new Scanner(System.in);
        int t = SC.nextInt();
        for(int i = 0 ; i<t ; i++){
            int n = SC.nextInt();
            int m = SC.nextInt();
            int r = SC.nextInt();
            int c = SC.nextInt();

            System.out.println(function(n,m,r,c));

        }

    }

    public static long function(int n , int m , int r , int c){
       long ans =(long) n*m - (long)(n-r) - (long)((r-1)*m + c) + (long)m*(n-r);
       return  ans;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that might cause the code to fail. The buggy code seems to have an issue with the calculation of the sum of Manhattan distances. Let's break down the problem and identify potential edge cases:

1. **Single Row or Column**: When there is only one row or one column, the movement of people will be linear, and the calculation should be straightforward.
2. **Person at the Edge**: When the person who leaves is at the edge of the grid (e.g., the first or last row or column), the movement of other people will be different.
3. **Person in the Middle**: When the person who leaves is in the middle of the grid, the movement will be more complex, involving both row and column shifts.
4. **Large Grid**: For a large grid, the calculation should still be correct, but the output might be very large, so we need to ensure the code handles large numbers correctly.

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

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

### Explanation:
1. **1 5 1 3**: A single row with 5 columns, and the person at position (1, 3) leaves. This tests the linear movement in a single row.
2. **5 1 3 1**: A single column with 5 rows, and the person at position (3, 1) leaves. This tests the linear movement in a single column.
3. **3 3 2 2**: A 3x3 grid, and the person at position (2, 2) leaves. This tests the movement in a small grid where the person is in the middle.
4. **10 10 5 5**: A larger grid (10x10), and the person at position (5, 5) leaves. This tests the movement in a larger grid where the person is in the middle.

This test input should help expose any issues in the buggy code, especially in the calculation of the sum of Manhattan distances.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sliding
time_limit_ms: 1000
memory_limit_mb: 256
Description: Red was ejected. They were not the imposter.

There are $n$ rows of $m$ people. Let the position in the $r$-th row and the $c$-th column be denoted by $(r, c)$. Number each person starting from $1$ in row-major order, i.e., the person numbered $(r-1)\cdot m+c$ is initially at $(r,c)$.

The person at $(r, c)$ decides to leave. To fill the gap, let the person who left be numbered $i$. Each person numbered $j>i$ will move to the position where the person numbered $j-1$ is initially at. The following diagram illustrates the case where $n=2$, $m=3$, $r=1$, and $c=2$.

![](CDN_BASE_URL/9b0b8e601446e3410296d7c9b1ff8763)

Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at $(r_0, c_0)$ and then moved to $(r_1, c_1)$, the Manhattan distance is $|r_0-r_1|+|c_0-c_1|$.

Input: The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases.

The only line of each testcase contains $4$ integers $n$, $m$, $r$, and $c$ ($1\le r\le n\le 10^6$, $1 \le c \le m \le 10^6$), where $n$ is the number of rows, $m$ is the number of columns, and $(r,c)$ is the position where the person who left is initially at.

Output: For each test case, output a single integer denoting the sum of the Manhattan distances.

Examples:
input:
4
2 3 1 2
2 2 2 1
1 1 1 1
1000000 1000000 1 1
output:
6
1
0
1999998000000

Note: For the first test case, the person numbered $2$ leaves, and the distances of the movements of the person numbered $3$, $4$, $5$, and $6$ are $1$, $3$, $1$, and $1$, respectively. So the answer is $1+3+1+1=6$.

For the second test case, the person numbered $3$ leaves, and the person numbered $4$ moves. The answer is $1$.

Buggy Code:
import java.util.Scanner;

public class A
{
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int T=sc.nextInt();
        while (T-->0)
        {
            int n=sc.nextInt(),m=sc.nextInt(),r=sc.nextInt(),c=sc.nextInt();
            System.out.println((long)n*m-((r-1)*m+c)+(long)(m-1)*(n-r));
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that might cause the formula in the buggy code to fail. The formula in the buggy code is:

```java
System.out.println((long)n*m-((r-1)*m+c)+(long)(m-1)*(n-r));
```

This formula is intended to calculate the sum of the Manhattan distances for each person's movement. However, it might not handle certain edge cases correctly. Let's analyze the formula and identify potential issues:

1. **Edge Case: Single Row or Single Column:**
   - If there is only one row (`n = 1`) or one column (`m = 1`), the movement pattern will be different, and the formula might not handle it correctly.
2. **Edge Case: Person at the Last Position:**
   - If the person at the last position in the grid leaves, there should be no movement, and the sum of distances should be zero.
3. **Edge Case: Person at the First Position:**
   - If the person at the first position in the grid leaves, all subsequent people will move, and the sum of distances should be maximized.

Let's create a test input that covers these edge cases:

```plaintext
4
1 1 1 1
1 10 1 10
10 1 10 1
10 10 10 10
```

Explanation:
1. **Case 1: (1, 1, 1, 1)**
   - Single person in a single row and column.
   - Expected output: 0 (no movement).

2. **Case 2: (1, 10, 1, 10)**
   - Single row with 10 columns, and the person at the last column leaves.
   - Expected output: 0 (no movement).

3. **Case 3: (10, 1, 10, 1)**
   - Single column with 10 rows, and the person at the last row leaves.
   - Expected output: 0 (no movement).

4. **Case 4: (10, 10, 10, 10)**
   - 10 rows and 10 columns, and the person at the last position leaves.
   - Expected output: 0 (no movement).

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Competition Memories
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kevin used to get into Rio's Memories, and in Rio's Memories, a series of contests was once held. Kevin remembers all the participants and all the contest problems from that time, but he has forgotten the specific rounds, the distribution of problems, and the exact rankings.

There are $$$ m $$$ problems in total, with the $$$ i $$$-th problem having a difficulty of $$$ b_i $$$. Let each contest consist of $$$ k $$$ problems, resulting in a total of $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. This means that you select exactly $$$ \lfloor \frac{m}{k} \rfloor \cdot k $$$ problems for the contests in any combination you want, with each problem being selected at most once, and the remaining $$$m\bmod k$$$ problems are left unused. For example, if $$$m = 17$$$ and $$$k = 3$$$, you should create exactly $$$5$$$ contests consisting of $$$3$$$ problems each, and exactly $$$2$$$ problems will be left unused.

There are $$$ n $$$ participants in the contests, with Kevin being the $$$1$$$-st participant. The $$$ i $$$-th participant has a rating of $$$ a_i $$$. During the contests, each participant solves all problems with a difficulty not exceeding their rating, meaning the $$$ i $$$-th participant solves the $$$ j $$$-th problem if and only if $$$ a_i \geq b_j $$$. In each contest, Kevin's rank is one plus the number of participants who solve more problems than he does.

For each $$$ k = 1, 2, \ldots, m $$$, Kevin wants to know the minimum sum of his ranks across all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. In other words, for some value of $$$k$$$, after selecting the problems for each contest, you calculate the rank of Kevin in each contest and sum up these ranks over all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. Your goal is to minimize this value.

Note that contests for different values of $$$k$$$ are independent. It means that for different values of $$$k$$$, you can select the distribution of problems into the contests independently.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$ t $$$ ($$$ 1 \le t \le 5\cdot 10^4 $$$).

The first line of each test case contains two integers $$$ n $$$ and $$$ m $$$ ($$$ 1 \le n, m \leq 3\cdot 10^5 $$$) — the number of participants and the number of problems.

The second line of each test case contains $$$ n $$$ integers $$$ a_1, a_2, \ldots, a_n $$$ ($$$ 0 \le a_i \le 10^9 $$$) — the rating of each participant.

The third line of each test case contains $$$ m $$$ integers $$$ b_1, b_2, \ldots, b_m $$$ ($$$ 0 \le b_i \le 10^9 $$$) — the difficulty of each problem.

It is guaranteed that both the sum of $$$ n $$$ and the sum of $$$ m $$$ over all test cases do not exceed $$$ 3 \cdot 10^5 $$$.

Output: For each test case, output $$$m$$$ integers — the minimum sum of Kevin's ranks for each $$$ k = 1, 2, \ldots, m$$$.

Examples:
input:
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output:
7 4 2 3
6 2 1 1 2
7 3 2 1 1 1 1
15 9 5 4 4 4

Note: For the first test case:

When $$$k=1$$$, since each contest only contains one problem, the distribution is in fact unique. For example, in the contest which only includes the third problem (which has a difficulty of $$$4$$$), all participants except the $$$2$$$-nd can solve it. Since no one solves strictly more problems than Kevin, his ranking in this contest is $$$1$$$. Similarly, in all $$$4$$$ contests, Kevin's rankings are $$$1,3,1,2$$$, and the sum is $$$7$$$.

When $$$k=2$$$, one optimal way is to choose the $$$1$$$-st and the $$$3$$$-rd problem to form a contest, while the $$$2$$$-nd and $$$4$$$-th for another. In the former contest, $$$4$$$ participants respectively solve $$$2,1,2,2$$$ problems, so Kevin's ranking is $$$1$$$; in the latter one, they respectively solve $$$0,0,2,1$$$, since there are $$$2$$$ participants ($$$3$$$-rd and $$$4$$$-th) solve more problems than Kevin, his ranking is $$$1+2=3$$$. Thus the answer is $$$1+3=4$$$. It can be proven that there's no way to achieve a lower sum.

When $$$k=3$$$, we can simply choose the $$$1$$$-st, the $$$3$$$-rd, and the $$$4$$$-th problem to make a contest, and Kevin has a ranking of $$$2$$$, which is optimal.

When $$$k=4$$$, since there's only one contest, the distribution is also unique, and Kevin's ranking is $$$3$$$.

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

int t;
int n, m;
vector<int> a;
vector<int> b;
vector<int> participant_solved[11]; // Max n <= 10
int problem_solved_by[11][11]; // Max n, m <= 10

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> t;
    while (t--) {
        cin >> n >> m;
        a.resize(n);
        b.resize(m);
        for (int i = 0; i < n; ++i)
            cin >> a[i];
        for (int i = 0; i < m; ++i)
            cin >> b[i];

        // Precompute which participant can solve which problem
        for (int i = 0; i < n; ++i) {
            participant_solved[i].clear();
            for (int j = 0; j < m; ++j) {
                if (a[i] >= b[j]) {
                    // Participant i can solve problem j
                    problem_solved_by[i][j] = 1;
                    participant_solved[i].push_back(j);
                } else {
                    problem_solved_by[i][j] = 0;
                }
            }
        }

        vector<int> kevin_solved;
        for (int j = 0; j < m; ++j) {
            if (problem_solved_by[0][j])
                kevin_solved.push_back(j);
        }
        int s = kevin_solved.size(); // Number of problems Kevin can solve
        for (int k = 1; k <= m; ++k) {
            int c = m / k; // Number of contests
            if (c == 0) {
                cout << n << " "; // If no contests, Kevin's rank sum is n (max rank)
                continue;
            }

            int min_rank_sum = n * c; // Max possible rank sum
            // Generate all possible combinations of problems into contests
            vector<int> indices(m);
            for (int i = 0; i < m; ++i)
                indices[i] = i;
            // Generate all possible subsets of size k * c from m problems
            vector<int> select_indices(m, 0);
            for (int i = 0; i < k * c; ++i)
                select_indices[i] = 1;
            int curr_min_rank_sum = min_rank_sum;
            do {
                vector<int> selected_problems;
                for (int i = 0; i < m; ++i) {
                    if (select_indices[i])
                        selected_problems.push_back(indices[i]);
                }
                sort(selected_problems.begin(), selected_problems.end());
                // Now generate all possible groupings of selected problems into c contests of size k
                vector<int> per(selected_problems.size());
                for (int i = 0; i < selected_problems.size(); ++i)
                    per[i] = i;
                do {
                    int rank_sum = 0;
                    for (int i = 0; i < c; ++i) {
                        vector<int> contest_problems;
                        for (int j = 0; j < k; ++j) {
                            contest_problems.push_back(selected_problems[per[i * k + j]]);
                        }
                        // Count number of problems Kevin solves in this contest
                        int kevin_solve = 0;
                        for (int p : contest_problems) {
                            if (problem_solved_by[0][p]) ++kevin_solve;
                        }
                        // For other participants
                        int better_participants = 0;
                        for (int p = 1; p < n; ++p) {
                            int participant_solve = 0;
                            for (int prob : contest_problems) {
                                if (problem_solved_by[p][prob])
                                    participant_solve++;
                            }
                            if (participant_solve > kevin_solve)
                                better_participants++;
                        }
                        rank_sum += 1 + better_participants;
                    }
                    if (rank_sum < curr_min_rank_sum)
                        curr_min_rank_sum = rank_sum;
                } while (next_permutation(per.begin(), per.end()));
            } while (prev_permutation(select_indices.begin(), select_indices.end()));
            cout << curr_min_rank_sum << " ";
        }
        cout << "\n";
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Competition Memories
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kevin used to get into Rio's Memories, and in Rio's Memories, a series of contests was once held. Kevin remembers all the participants and all the contest problems from that time, but he has forgotten the specific rounds, the distribution of problems, and the exact rankings.

There are $$$ m $$$ problems in total, with the $$$ i $$$-th problem having a difficulty of $$$ b_i $$$. Let each contest consist of $$$ k $$$ problems, resulting in a total of $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. This means that you select exactly $$$ \lfloor \frac{m}{k} \rfloor \cdot k $$$ problems for the contests in any combination you want, with each problem being selected at most once, and the remaining $$$m\bmod k$$$ problems are left unused. For example, if $$$m = 17$$$ and $$$k = 3$$$, you should create exactly $$$5$$$ contests consisting of $$$3$$$ problems each, and exactly $$$2$$$ problems will be left unused.

There are $$$ n $$$ participants in the contests, with Kevin being the $$$1$$$-st participant. The $$$ i $$$-th participant has a rating of $$$ a_i $$$. During the contests, each participant solves all problems with a difficulty not exceeding their rating, meaning the $$$ i $$$-th participant solves the $$$ j $$$-th problem if and only if $$$ a_i \geq b_j $$$. In each contest, Kevin's rank is one plus the number of participants who solve more problems than he does.

For each $$$ k = 1, 2, \ldots, m $$$, Kevin wants to know the minimum sum of his ranks across all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. In other words, for some value of $$$k$$$, after selecting the problems for each contest, you calculate the rank of Kevin in each contest and sum up these ranks over all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. Your goal is to minimize this value.

Note that contests for different values of $$$k$$$ are independent. It means that for different values of $$$k$$$, you can select the distribution of problems into the contests independently.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$ t $$$ ($$$ 1 \le t \le 5\cdot 10^4 $$$).

The first line of each test case contains two integers $$$ n $$$ and $$$ m $$$ ($$$ 1 \le n, m \leq 3\cdot 10^5 $$$) — the number of participants and the number of problems.

The second line of each test case contains $$$ n $$$ integers $$$ a_1, a_2, \ldots, a_n $$$ ($$$ 0 \le a_i \le 10^9 $$$) — the rating of each participant.

The third line of each test case contains $$$ m $$$ integers $$$ b_1, b_2, \ldots, b_m $$$ ($$$ 0 \le b_i \le 10^9 $$$) — the difficulty of each problem.

It is guaranteed that both the sum of $$$ n $$$ and the sum of $$$ m $$$ over all test cases do not exceed $$$ 3 \cdot 10^5 $$$.

Output: For each test case, output $$$m$$$ integers — the minimum sum of Kevin's ranks for each $$$ k = 1, 2, \ldots, m$$$.

Examples:
input:
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output:
7 4 2 3
6 2 1 1 2
7 3 2 1 1 1 1
15 9 5 4 4 4

Note: For the first test case:

When $$$k=1$$$, since each contest only contains one problem, the distribution is in fact unique. For example, in the contest which only includes the third problem (which has a difficulty of $$$4$$$), all participants except the $$$2$$$-nd can solve it. Since no one solves strictly more problems than Kevin, his ranking in this contest is $$$1$$$. Similarly, in all $$$4$$$ contests, Kevin's rankings are $$$1,3,1,2$$$, and the sum is $$$7$$$.

When $$$k=2$$$, one optimal way is to choose the $$$1$$$-st and the $$$3$$$-rd problem to form a contest, while the $$$2$$$-nd and $$$4$$$-th for another. In the former contest, $$$4$$$ participants respectively solve $$$2,1,2,2$$$ problems, so Kevin's ranking is $$$1$$$; in the latter one, they respectively solve $$$0,0,2,1$$$, since there are $$$2$$$ participants ($$$3$$$-rd and $$$4$$$-th) solve more problems than Kevin, his ranking is $$$1+2=3$$$. Thus the answer is $$$1+3=4$$$. It can be proven that there's no way to achieve a lower sum.

When $$$k=3$$$, we can simply choose the $$$1$$$-st, the $$$3$$$-rd, and the $$$4$$$-th problem to make a contest, and Kevin has a ranking of $$$2$$$, which is optimal.

When $$$k=4$$$, since there's only one contest, the distribution is also unique, and Kevin's ranking is $$$3$$$.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<cmath>
#include<bitset>
#include<cstring>
#include<queue>
#define pb push_back
#define fi first
#define se second
using namespace std;
using ll=long long;
typedef pair<int,int>pii;
const int N=3e5+10;
void Silverwolf(){
    int n,m;
    cin>>n>>m;//n个人m个问题
    vector<int>a(n+1);
    for(int i=1;i<=n;i++)cin>>a[i];
    vector<int>b(m+1);
    for(int i=1;i<=m;i++)cin>>b[i];
    if(n==1){
        for(int i=1;i<=m;i++)cout<<1<<' ';cout<<'\n';
        return;
    }
    sort(b.begin()+1,b.end());
    sort(a.begin()+2,a.end());
    vector<int>rk(m+1,0);//每一题的得分
    for(int i=1;i<=m;i++){
        if(b[i]<=a[1])rk[i]=1;
        else{
            int l=2,r=n;
            //如果没有人做的出来
            if(a[n]<b[i]){
                rk[i]=1;
            }else{
                while(l<r){
                    int mid=l+r>>1;
                    if(a[mid]>=b[i])r=mid;
                    else l=mid+1;
                }
                rk[i]=n-l+2;
            }
        }
    }
    // for(int i=1;i<=m;i++)cout<<b[i]<<' ';cout<<'\n';
    // for(int i=1;i<=m;i++)cout<<rk[i]<<' ';cout<<'\n';
    sort(rk.begin()+1,rk.end());
    for(int i=1;i<=m;i++){
        ll ans=0;
        for(int j=i;j<=m;j+=i)ans+=rk[j];
        cout<<ans<<' ';
    }cout<<'\n';
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;cin>>T;while(T--)
    Silverwolf();
    //愿艾利欧三度为你窥见，令你的生命永不停歇，骇入永远成功，游戏永远胜利。
    //游戏就只是为了游戏，仅此而已
    //acm这种东西，三两下搞定就好啦
    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 6
10 5 1
1 2 3 4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Competition Memories
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kevin used to get into Rio's Memories, and in Rio's Memories, a series of contests was once held. Kevin remembers all the participants and all the contest problems from that time, but he has forgotten the specific rounds, the distribution of problems, and the exact rankings.

There are $$$ m $$$ problems in total, with the $$$ i $$$-th problem having a difficulty of $$$ b_i $$$. Let each contest consist of $$$ k $$$ problems, resulting in a total of $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. This means that you select exactly $$$ \lfloor \frac{m}{k} \rfloor \cdot k $$$ problems for the contests in any combination you want, with each problem being selected at most once, and the remaining $$$m\bmod k$$$ problems are left unused. For example, if $$$m = 17$$$ and $$$k = 3$$$, you should create exactly $$$5$$$ contests consisting of $$$3$$$ problems each, and exactly $$$2$$$ problems will be left unused.

There are $$$ n $$$ participants in the contests, with Kevin being the $$$1$$$-st participant. The $$$ i $$$-th participant has a rating of $$$ a_i $$$. During the contests, each participant solves all problems with a difficulty not exceeding their rating, meaning the $$$ i $$$-th participant solves the $$$ j $$$-th problem if and only if $$$ a_i \geq b_j $$$. In each contest, Kevin's rank is one plus the number of participants who solve more problems than he does.

For each $$$ k = 1, 2, \ldots, m $$$, Kevin wants to know the minimum sum of his ranks across all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. In other words, for some value of $$$k$$$, after selecting the problems for each contest, you calculate the rank of Kevin in each contest and sum up these ranks over all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. Your goal is to minimize this value.

Note that contests for different values of $$$k$$$ are independent. It means that for different values of $$$k$$$, you can select the distribution of problems into the contests independently.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$ t $$$ ($$$ 1 \le t \le 5\cdot 10^4 $$$).

The first line of each test case contains two integers $$$ n $$$ and $$$ m $$$ ($$$ 1 \le n, m \leq 3\cdot 10^5 $$$) — the number of participants and the number of problems.

The second line of each test case contains $$$ n $$$ integers $$$ a_1, a_2, \ldots, a_n $$$ ($$$ 0 \le a_i \le 10^9 $$$) — the rating of each participant.

The third line of each test case contains $$$ m $$$ integers $$$ b_1, b_2, \ldots, b_m $$$ ($$$ 0 \le b_i \le 10^9 $$$) — the difficulty of each problem.

It is guaranteed that both the sum of $$$ n $$$ and the sum of $$$ m $$$ over all test cases do not exceed $$$ 3 \cdot 10^5 $$$.

Output: For each test case, output $$$m$$$ integers — the minimum sum of Kevin's ranks for each $$$ k = 1, 2, \ldots, m$$$.

Examples:
input:
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output:
7 4 2 3
6 2 1 1 2
7 3 2 1 1 1 1
15 9 5 4 4 4

Note: For the first test case:

When $$$k=1$$$, since each contest only contains one problem, the distribution is in fact unique. For example, in the contest which only includes the third problem (which has a difficulty of $$$4$$$), all participants except the $$$2$$$-nd can solve it. Since no one solves strictly more problems than Kevin, his ranking in this contest is $$$1$$$. Similarly, in all $$$4$$$ contests, Kevin's rankings are $$$1,3,1,2$$$, and the sum is $$$7$$$.

When $$$k=2$$$, one optimal way is to choose the $$$1$$$-st and the $$$3$$$-rd problem to form a contest, while the $$$2$$$-nd and $$$4$$$-th for another. In the former contest, $$$4$$$ participants respectively solve $$$2,1,2,2$$$ problems, so Kevin's ranking is $$$1$$$; in the latter one, they respectively solve $$$0,0,2,1$$$, since there are $$$2$$$ participants ($$$3$$$-rd and $$$4$$$-th) solve more problems than Kevin, his ranking is $$$1+2=3$$$. Thus the answer is $$$1+3=4$$$. It can be proven that there's no way to achieve a lower sum.

When $$$k=3$$$, we can simply choose the $$$1$$$-st, the $$$3$$$-rd, and the $$$4$$$-th problem to make a contest, and Kevin has a ranking of $$$2$$$, which is optimal.

When $$$k=4$$$, since there's only one contest, the distribution is also unique, and Kevin's ranking is $$$3$$$.

Buggy Code:
#include <iostream>
#include <fstream>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;

#define endl '\n'
#define int long long
#define fi first
#define se second
#define lc lson
#define rc rson
#define PII pair<int,pair<int,int>>

const int N = 2e5 + 10, M = 3050;
const int inf = 0x7f7f7f7f7f7f7f7f;
const int MOD = 998244353;

int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int qmi(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
        if (b & 1) res = res * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return res % MOD;
}
int lcm(int a, int b) { return a * b / gcd(a, b); }
int C(int a, int b, int p) {
    if (b > a) return 0;
    int res = 1;
    for (int i = 1, j = a; i <= b; ++i, --j) {
        res = res * j % p;
        res = res * qmi(i, p - 2) % p;
    }
    return res;
}
int lucas(int a, int b, int p) {               // 注意LL参数类型
    if (a < p && b < p) return C(a, b, p);
    return C(a % p, b % p, p) * lucas(a / p, b / p, p) % p;     // 递归让其到p范围内求解
}
int isqrt_newton(int n) {
    int x = 1;
    bool decreased = false;
    for (;;) {
        int nx = (x + n / x) >> 1;
        if (x == nx || (nx > x && decreased)) break;
        decreased = nx < x;
        x = nx;
    }
    return x;
}

void solve() {
    int n, m;
    cin >> n >> m;
    vector<int>a(n + 1), b(m + 1);
    vector<pair<int, int>>p(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        p[i] = { a[i],i };
    }
    for (int i = 1; i <= m; i++) {cin >> b[i];}

    sort(p.begin(), p.end());
    sort(b.begin(), b.end());
    sort(a.begin(), a.end());
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        if (p[i].second == 1) {
            cnt = i;
        }
    }
    int num = 0;
    for (int i = 1; i <= m; i++) {
        if (b[i] > p[cnt].first) {
            num = i;
            break;
        }
    }

    vector<int>cha(m + 1);
    for (int i = 1; i <= m; i++) {
        int k = lower_bound(a.begin(), a.end(), b[i]) - a.begin();
        if (k<0 || k>n) {
            break;
        }
        cha[i] = n - k + 1;
    }
    //cout << "---------" << endl;
    //for (int i = 1; i <= n; i++) {
    //    cout << a[i] << " ";
    //}
    //cout << endl;
    //for (int i = 1; i <= m; i++) {
    //    cout << b[i] << " ";
    //}
    //cout << endl;
    //for (int i = 1; i <= m; i++) {
    //    cout << cha[i] << " ";
    //}
    //cout << endl;
    //cout << "cnt:" << cnt << " " << "num:" << num << endl;
    //cout << "---------" << endl;

    for (int i = 1; i <= m; i++) {
        int ans = 0;
        int mn = inf;
        int y = m % i;
        for (int j = 1; j <= m; j += i) {
            //cout << "j:" << j << " " << "cha:" << cha[j] << endl;
            if ((j + i - 1) < num) {
                ans += 1;
            }
            else if (j > num) {
                ans += cha[j] + 1;
            }
            else {
                //cout << "y:" << y << endl;
                if (num + y <= m) {
                    ans += cha[num + y] + 1;
                }
                j = num + y;
            }
        }
        cout << ans << " " ;
    }
    cout << endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that can maximize the failure probability. The code seems to have issues with how it handles the distribution of problems and the calculation of ranks. Let's consider a small-scale test case that can highlight these issues.

### Key Points to Consider:
1. **Small Number of Participants and Problems**: This helps in verifying the logic step-by-step.
2. **Mixed Ratings and Difficulties**: Ensure that the ratings and difficulties are mixed to test the sorting and comparison logic.
3. **Special Cases for \( k \)**: Test \( k \) values that are edge cases, such as \( k = 1 \), \( k = m \), and \( k \) values that divide \( m \) evenly.

### Test Case:
Let's create a test case with:
- \( n = 4 \)
- \( m = 6 \)
- Mixed ratings and difficulties

### Test Input:
```plaintext
1
4 6
4 3 7 5
2 5 4 6 1 3
```

### Explanation:
- **Participants' Ratings**: \( a = [4, 3, 7, 5] \)
- **Problems' Difficulties**: \( b = [2, 5, 4, 6, 1, 3] \)

### Expected Output:
- For \( k = 1 \): Each problem is a separate contest. Kevin's ranks will be \( [1, 3, 1, 2, 1, 2] \). Sum = 10.
- For \( k = 2 \): Contests can be formed as \( [2, 5], [4, 6], [1, 3] \). Kevin's ranks will be \( [2, 3, 1] \). Sum = 6.
- For \( k = 3 \): Contests can be formed as \( [2, 5, 4], [6, 1, 3] \). Kevin's ranks will be \( [2, 1] \). Sum = 3.
- For \( k = 4 \): Contests can be formed as \( [2, 5, 4, 6] \). Kevin's rank will be 3. Sum = 3.
- For \( k = 5 \): Contests can be formed as \( [2, 5, 4, 6, 1] \). Kevin's rank will be 2. Sum = 2.
- For \( k = 6 \): Contests can be formed as \( [2, 5, 4, 6, 1, 3] \). Kevin's rank will be 3. Sum = 3.

### Final Test Input:
```plaintext
1
4 6
4 3 7 5
2 5 4 6 1 3
```

This test case should help expose any issues in the code's logic for handling the distribution of problems and the calculation of ranks.