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;
using ll=long long;
constexpr ll mod=998244353;
ll A,B,C,D;
ll DP[3001][3001][2][2];

int main(){
	cin>>A>>B>>C>>D;
	DP[A][B][1][1]=1;
	for(int i=A;i<=C;i++){
		for(int j=B;j<=D;j++){
			DP[i][j][0][0]=(DP[i-1][j-1][0][0]+DP[i-1][j-1][0][1]+DP[i-1][j-1][1][0]+DP[i-1][j-1][1][1])*(i-1)*(j-1)%mod;
			DP[i][j][1][0]=(DP[i-1][j][0][0]+DP[i-1][j][0][1]+(DP[i-1][j][1][0]+DP[i-1][j][1][1])*j)%mod;
			DP[i][j][0][1]=(DP[i][j-1][0][0]+DP[i][j-1][1][0]+(DP[i][j-1][0][1]+DP[i][j-1][1][1])*i)%mod;
		}
	}
	cout<<(DP[C][D][0][0]+DP[C][D][0][1]+DP[C][D][1][0]+DP[C][D][1][1])%mod<<endl;
}