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>
using namespace std;
long long a,b,c,d,e,i,j,ii,jj,zx,xc,mod=998244353LL,dp[3009][3009];
int main(){
	ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>a>>b>>c>>d;
	dp[a][b]=1;
	for(i=a; i<=c; i++){
		for(j=b; j<=d; j++){
		    if(i==a&&j==b) continue;
		    if(i>a&&j>b){
			dp[i][j]=dp[i-1][j]*j+dp[i][j-1]*i;
			zx=dp[i-1][j-1]*(i-1)*(j-1);zx%=mod;
			dp[i][j]=(dp[i][j]-zx+mod)%mod;
		    }else{
		        dp[i][j]=dp[i-1][j]*j+dp[i][j-1]*i;
		        dp[i][j]%=mod;
		    }
		}
	}
	cout<<dp[c][d];
	return 0;
}