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;

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int a,b,c,d;
	cin >> a >> b >> c >> d;
	long long mod=998244353;
	
	vector<vector<long long>> dp(c+1,vector<long long>(d+1));
	dp[a][b]=1;
	for(int i=a;i<=c;i++){
		for(int j=b;j<=d;j++){
			if(i==a && j==b) continue;
			if(i==a){
				dp[i][j]=dp[i][j-1]*i%mod;
			}else if(j==b){
				dp[i][j]=dp[i-1][j]*j%mod;
			}else{
				dp[i][j]=(dp[i-1][j]*j%mod+dp[i][j-1]*i%mod+mod-(i-1)*(j-1)%mod*dp[i-1][j-1]%mod)%mod;
			}
		}
	}
	cout << dp[c][d] << "\n";
	return 0;
}