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;
typedef long long ll;
const ll mod=998244353;
const int N=3005;
int n;
ll dp[N][N];
int a,b,c,d;
int main(){
	ios::sync_with_stdio(false);
	cin >> a >> b >> c >> d;
	for(int i=a; i<=c ;i++){
		for(int j=b; j<=d ;j++){
			if(i==a && j==b){
				dp[i][j]=1;
				continue;
			}
			dp[i][j]=dp[i-1][j]*j+dp[i][j-1]*i-dp[i-1][j-1]*((i-1)*(j-1));
			dp[i][j]=(dp[i][j]%mod+mod)%mod;
		}
	}
	cout << dp[c][d] << '\n';
}