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<cstdio>
#include<cstring>
#include<iostream>
const int p=998244353;
using namespace std;
int T,n,a,b,c,d;long long dp[3200][3200];
int main(){
	scanf("%d%d%d%d",&a,&b,&c,&d);
	dp[a][b]=1;
	
	for(int i=a;i<=c;i++)
	for(int j=b;j<=d;j++)
	dp[i+1][j]=(dp[i+1][j]+dp[i][j]*j%p)%p,dp[i][j+1]=(dp[i][j+1]+dp[i][j]*i%p)%p,
	dp[i+1][j+1]=(dp[i+1][j+1]+p-dp[i][j]*i%p*j%p)%p;
	
	printf("%d",dp[c][d]);
	
}