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 <stdio.h>
#define MOD 998244353
long long int DP[3010][3010];
int main()
{
	int a,b,c,d;
	scanf("%d%d%d%d",&a,&b,&c,&d);
	
	long long int C = 1;
	for(int j=c;j>=a;j--)
	{
		DP[d][j] = C;
		C*=d;
		C%=MOD;
	}
	
	for(int i=d-1;i>=b;i--)
	{
		C = 0;
		for(int j=c;j>=a;j--)
		{
			DP[i][j] = (C + j*DP[i+1][j])%MOD;
			
			//C *= i;
			C += DP[i+1][j];
			C *= i;
			C %= MOD;
			//printf("%d %d %lld!!\n",i,j,DP[i][j]);
		}
	}
	
	printf("%lld",DP[b][a]);
}