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>
#define mod 998244353
using namespace std;
long long a,b,c,d,dp[3005][3005],k,i;
int main(){
	cin>>a>>b>>c>>d;
	dp[a+1][b]=b; dp[a][b+1]=a;
dp[a][b]=1;
    for(k=a+1;k<=c;k++)
    dp[k][b]=dp[k-1][b]*b%mod;
    
    for(k=b+1;k<=d;k++)
    dp[a][k]=dp[a][k-1]*a%mod;
	for(int k=a+1;k<=c;k++)
	for(int i=b+1;i<=d;i++)
	{
		dp[k][i]=(dp[k-1][i]*i%mod+dp[k][i-1]*k%mod-dp[k-1][i-1]*(k-1)%mod*(i-1)%mod+mod)%mod;
	}
	cout<<dp[c][d];
}