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 ll long long


using namespace std;

ll mod=998244353;
ll dp[3001][3001];
int main(){
	int a,b;cin>>a>>b;
	int c,d;cin>>c>>d;
	dp[a][b]=1;
	for(int i=a+1;i<=c;i++){
		dp[i][b]=dp[i-1][b]*b;
		dp[i][b]%=mod;
	}
	for(int j=b+1;j<=d;j++){
		dp[a][j]=dp[a][j-1]*a;
		dp[a][j]%=mod;
	}
	for(int i=a+1;i<=c;i++){
		for(int j=b+1;j<=d;j++){
			dp[i][j]=(dp[i-1][j]*(j%mod)+dp[i][j-1]*(i%mod))%mod;
			dp[i][j]-=(dp[i-1][j-1]*((i-1)%mod)*((j-1))%mod);
			dp[i][j]+=mod;
		
			dp[i][j]%=mod;
		}
	}

	cout<<dp[c][d];
}