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 mo 998244353
#define ll long long
using namespace std;
ll A,B,C,D,ans,dp[3010][3010][2];
void add(ll &x,ll y){x=(x+y)%mo;}
int main(){
	cin>>A>>B>>C>>D;
	dp[A][B][0]=1;
	for (ll i=A;i<=C;i++){
		for (ll j=B;j<=D;j++){
			add(dp[i][j+1][1],(dp[i][j][1]+dp[i][j][0])*i);
			add(dp[i+1][j][0],dp[i][j][0]*j+dp[i][j][1]);
		}
	}
	cout<<(dp[C][D][0]+dp[C][D][1])%mo<<endl;
	return 0;
}