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<bits/stdc++.h>
const int N=3005,mo=998244353;
int f[N][N][2],A,B,C,D,i,j;
int main(){
	scanf("%d%d%d%d",&A,&B,&C,&D);
	f[A][B][0]=1;
	for(i=A;i<=C;++i)for(j=B;j<=D;++j){
		f[i+1][j][0]=(f[i+1][j][0]+1ll*f[i][j][0]*j)%mo;
		f[i][j+1][1]=(f[i][j+1][1]+1ll*f[i][j][0]*i)%mo;
		f[i+1][j][0]=(f[i+1][j][0]+f[i][j][1])%mo;
		f[i][j+1][1]=(f[i][j+1][1]+1ll*f[i][j][1]*i)%mo;
	}
	printf("%d\n",(f[C][D][0]+f[C][D][1])%mo);
}