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;
 
int main(){
	long long int A,B,C,D; 
	cin >> A >> B >> C >> D;
    long long int x[C+2][D+2];
    x[C][D]=1;
  	for(int j=1;j<D+2;j++){
    x[C+1][j]=0;
    }
    for(int i=1;i<C+2;i++){
    x[i][D+1]=0;
    }
    for(int j=D-1;j>0;j--){
    x[C][j]=(C*x[C][j+1]+j*x[C+1][j]+998244353-(C*j*x[C+1][j+1] %  998244353)) % 998244353;
    }
    for(int i=C-1;i>0;i--){
	    for(int j=D;j>0;j--){
    	x[i][j]=(i*x[i][j+1]+j*x[i+1][j]+998244353-(i*j*x[i+1][j+1] %  998244353)) %  998244353;
    	}
    }
    	cout << x[A][B] << endl;
}