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 rep(i,n) for(int i=0; i<(int)(n); i++)

using namespace std;
using LL = long long;
using P = pair<int,int>;
const LL mod=998244353;
const int Max_N=3010;

LL dp[Max_N][Max_N];

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

	return 0;
}