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: Wrong Answer)

#include <stdio.h>

#define Mod 998244353

int main()
{
	int A, B, C, D;
	scanf("%d %d %d %d", &A, &B, &C, &D);
	if (A == C && B == D) {
		printf("1\n");
		fflush(stdout);
		return 0;
	}
	
	int i, j;
	long long dp[2][3002] = {};
	dp[0][0] = 1;
	for (j = 1; j <= D - B; j++) dp[0][j] = dp[0][j-1] * A % Mod;
	for (i = 1; i <= C - A; i++) {
		dp[i%2][0] = dp[1-i%2][0] * B % Mod;
		for (j = 1; j <= D - B; j++) dp[i%2][j] = (dp[i%2][j-1] * (A + i) + dp[1-i%2][j] * (B + j) - dp[1-i%2][j-1] * (A + i - 1) * (B + j - 1)) % Mod;
	}
	
	
	printf("%lld\n", dp[(C-A)%2][D-B]);
	fflush(stdout);
	return 0;
}