Problem: 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.

Examples

Input

1 1 2 2


Output

3


Input

2 1 3 4


Output

65


Input

31 41 59 265


Output

387222020

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int mod=998244353;
ll f[3005][3005];
int A,B,C,D;
int main() {
	scanf("%d%d%d%d",&A,&B,&C,&D);
	f[A][B]=1;
	for(int i=A;i<=C;i++) {
		for(int p=B;p<=D;p++) {
			if(i>A) {
				f[i][p]+=f[i-1][p]*p;
			}
			if(p>B) {
				f[i][p]+=f[i][p-1]*i;
			}
			if(i>A && p>B) {
				f[i][p]-=f[i-1][p-1]*(i-1)%mod*(p-1);
			}
			f[i][p]=(f[i][p]%mod+mod)%mod;
		}
	}
	printf("%lld\n",f[C][D]);
	return 0;
}