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;
typedef long long intl;
intl dp[3010][3010];
intl MOD= 998244353;
int main(){
	intl a,b,c,d;
	cin>>a>>b>>c>>d;
	intl n=c-a;
	intl m=d-b;
	for(intl i=0;i<=n;i++){
		for(intl j=0;j<=m;j++){
			if(i==0&& j==0){
				dp[i][j]=1;
			}
			else if(i==0){
				dp[i][j]=dp[i][j-1]*(i+a)%MOD;	
			}
			else if(j==0){
				dp[i][j]=dp[i-1][j]*(j+b)%MOD;
			}
			else{
				dp[i][j]=(dp[i-1][j]*(j+b)+dp[i][j-1]*(i+a)-(dp[i-1][j-1]*(j+b-1)*(i+a-1))%MOD+MOD)%MOD;
			}
			
		}
	}
	
	
	cout<<dp[n][m]<<endl;
}