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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 998244353;
int a,b,c,d;
ll mem[3005][3005];
ll dp(int x, int y){
	if (x == a && y == b) return 1ll;
	if (mem[x][y] != -1) return mem[x][y];
	ll res = 0;
	if (x > a){
		res += y*dp(x-1,y);
	}
	if (y > b){
		res += x*dp(x,y-1);
	}
	if (x > a && y > b) res -= (x-1)*(y-1)*dp(x-1,y-1);
	res = ((res % mod) + mod )%mod;
	return mem[x][y] = res;
}

int main(){
	scanf("%d%d%d%d",&a,&b,&c,&d);
	memset(mem,-1,sizeof(mem));
	printf("%lld",dp(c,d));
	
}