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;
const long long MOD=998244353;
int A,B,C,D;
long long dp[3003][3003];
int main(){
	cin>>A>>B>>C>>D;
	dp[A][B]=1;
	for(int i=B+1;i<=D;i++)dp[A][i]=dp[A][i-1]*A%MOD;
	for(int i=A+1;i<=C;i++)dp[i][B]=dp[i-1][B]*B%MOD;
	for(int i=A+1;i<=C;i++)
		for(int j=B+1;j<=D;j++){
			dp[i][j]=(dp[i][j-1]*i%MOD+dp[i-1][j]*j%MOD-dp[i-1][j-1]*(i-1)*(j-1)%MOD+MOD*20)%MOD;
	//		cout<<i<<" "<<j<<" "<<dp[i][j]<<"\n";
		}
	//for(int i=1;i<=C;i++,puts(""))for(int j=1;j<=D;j++)cout<<dp[i][j]<<" ";
	cout<<dp[C][D];
}