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>
using namespace std;
#define int long long
int dp[5005][5005];
const int mod=998244353;
int32_t main(){
	int a, b, c, d; cin>>a>>b>>c>>d;
	dp[a][b]=1;
	for(int i=a; i<=c; i++) for(int j=b; j<=d; j++) if(!(i==a && j==b)){
		dp[i][j]=(dp[i-1][j]*j+dp[i][j-1]*i+mod-dp[i-1][j-1]*(i-1)*(j-1))%mod;
      	if(dp[i][j]<0) dp[i][j]+=mod;
    }
	cout<<dp[c][d]<<endl;
	return 0;
}