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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

const long long M=998244353;
long long f[3005][3005];
int a,b,c,d;

int main () {
	
	scanf ("%d%d%d%d",&a,&b,&c,&d);
	for (int i=a;i<=c;i++) {
		for (int j=b;j<=d;j++) {
			if (i==a&&j==b) {f[i][j]=1; continue;}
			f[i][j]=(f[i-1][j]*j+f[i][j-1]*i-f[i-1][j-1]*(i-1)*(j-1))%M;
			if (f[i][j]<0) f[i][j]+=M;
		}
	}
	printf ("%lld\n",f[c][d]);
	
	return 0;
}