Problem:
Score : 600 points

Problem Statement

We have a grid with A horizontal rows and B vertical columns, with the squares painted white. On this grid, we will repeatedly apply the following operation:

Assume that the grid currently has a horizontal rows and b vertical columns. Choose "vertical" or "horizontal".

If we choose "vertical", insert one row at the top of the grid, resulting in an (a+1) \times b grid.

If we choose "horizontal", insert one column at the right end of the grid, resulting in an a \times (b+1) grid.

Then, paint one of the added squares black, and the other squares white.

Assume the grid eventually has C horizontal rows and D vertical columns. Find the number of ways in which the squares can be painted in the end, modulo 998244353.

Constraints

1 \leq A \leq C \leq 3000

1 \leq B \leq D \leq 3000

A, B, C, and D are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the number of ways in which the squares can be painted in the end, modulo 998244353.

Sample Input 1

1 1 2 2

Sample Output 1

3

Any two of the three squares other than the bottom-left square can be painted black.

Sample Input 2

2 1 3 4

Sample Output 2

65

Sample Input 3

31 41 59 265

Sample Output 3

387222020

Solution in C (GCC 9.2.1) (judge verdict: Runtime Error)

#include <stdio.h>

#define MAX_C 10001
#define MAX_D 10001
#define MOD_NUM 998244353

int main(void)
{
    int A, B, C, D;
    int MAX_NUM;
    int dp[MAX_C][MAX_D];
    scanf("%d %d %d %d", &A, &B, &C, &D);

    //dp[A][B] = 1;
    for (int i = 0; i <= A; i++) {
        for (int j = 0; j <= B; j++) {
            dp[i][j] = 1;
        }
    }

    for (int i = A + 1;i <= C; i++) {
        dp[i][B] = dp[i - 1][B] * B;
    }
    for (int i = B + 1;i <= D; i++) {
        dp[A][i] = dp[A][i - 1] * A;
    }
    
    int counter = 1;
    for (int i = A + 1; i <= C; i++) {
        for (int j = B + 1; j <= D; j++) {
            dp[i][j] = dp[i - 1][j] * j + dp[i][j - 1] * i - (i - 1) * (j - 1) * dp[i - 1][j - 1];
            counter += 1;
        }
    }

    printf("%d", dp[C][D] % MOD_NUM);


    return 0;
}