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<iostream>
using namespace std;
const long long mod=998244353;
long long a,b,n,m;
long long f[3005][3005];
int main(){
	cin>>a>>b>>n>>m;
	f[a][b]=1;
	for(long long i=a;i<=n;i++)
	for(long long j=b;j<=m;j++){
		if(i==a&&j==b) continue;
		if(b==j) {
			f[i][j]=f[i-1][j]*j%mod;
			continue;
		}
		if(a==i){
			f[i][j]=i*f[i][j-1]%mod;
			continue;
		}
		f[i][j]=(long long)j*f[i-1][j]+(long long)i*f[i][j-1]-(long long)(j-1)*(i-1)*f[i-1][j-1]%mod;
		f[i][j]=(long long)f[i][j]%mod;
	}
/*	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			cout<<f[i][j]<<' ';
		}
		cout<<endl;
	}*/
	cout<<f[n][m]<<endl;
	return 0;
	
}