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>
typedef int LL;
const LL maxn=3e3+9,mod=998244353;
LL a,b,c,d;
LL f[maxn][maxn];
LL add(LL x,LL y){ return x+=y,x>=mod?x-mod:x; }
LL mul(LL x,LL y){ return 1ll*x*y%mod; }
LL dec(LL x,LL y){ return x-=y,x<0?x+mod:x; }
int main(){
	scanf("%d%d%d%d",&a,&b,&c,&d);
	f[a][b]=1;
	for(LL i=a;i<=c;++i){
		for(LL j=b;j<=d;++j){
			if(i==a && j==b) continue;
			if(i==a){
				f[i][j]=mul(f[i][j-1],i);
			}else if(j==b){
				f[i][j]=mul(f[i-1][j],j);
			}else{
				f[i][j]=add(mul(f[i-1][j],j),mul(f[i][j-1],i));
				f[i][j]=dec(f[i][j],mul(f[i-1][j-1],1ll*(i-1)*(j-1)%mod));
			}
		}
	}
	printf("%d\n",f[c][d]);
	return 0;
}