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:
//A
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const long long mod=998244353;
long long dp[3030][3030];
int main()
{
	long long a,b,c,d;
	scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
	dp[a][b]=1;
	for(int i=b+1;i<=d;i++)
		dp[a][i]=(dp[a][i-1]*a%mod)%mod;
	for(int i=a+1;i<=c;i++){
		dp[i][b]=(dp[i-1][b]*b%mod)%mod;
		for(int j=b+1;j<=d;j++){
			dp[i][j]=((dp[i-1][j]*j)%mod+(dp[i][j-1]*i%mod)%mod-(dp[i-1][j-1]*(((i-1)*(j-1))%mod))%mod+mod)%mod;
		}
	}
	printf("%lld",dp[c][d]);
}