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:
#pragma GCC optimize("Ofast")
#include<stdio.h>
#include<algorithm>
using namespace std;
long long dp[3005][3005],mod=998244353;
int main(){
	long long i,j,a,b,c,d;
	scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
	for(i=a;i<=c;i++){
		for(j=b;j<=d;j++){
			if(i==a&&j==b)
				dp[i][j]=1;
			else if(i==a)
				dp[i][j]=dp[i][j-1]*i%mod;
			else if(j==b)
				dp[i][j]=dp[i-1][j]*j%mod;
			else
				dp[i][j]=(dp[i-1][j]*j%mod+dp[i][j-1]*i%mod-dp[i-1][j-1]*(i-1)%mod*(j-1)%mod+mod)%mod;
		}
	}
	printf("%lld\n",dp[c][d]);
}