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<stdio.h>
#include<iostream>
#define it register int
#define ct const int
#define il inline
using namespace std;
typedef long long ll;
#define rll register ll
#define cll const ll
#define P 998244353
const int N=5005;
int f[N][N],a,b,c,d;
int main(){ 
	scanf("%d%d%d%d",&a,&b,&c,&d);
	f[a][b]=1;
	for(it i=a,j;i<=c;++i)
		for(j=b;j<=d;++j){
			if(i==a&&j==b) continue;
			f[i][j]=(f[i][j]+(0ll+i)*f[i][j-1]+(0ll+j)*f[i-1][j]-(i-1ll)*(j-1ll)%P*f[i-1][j-1])%P,f[i][j]=(f[i][j]<0?f[i][j]+P:f[i][j]);
		}
	printf("%d",f[c][d]);
	return 0;
}