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 int long long
#define iris 998244353
using namespace std;

int dp[3010][3010];

int sana(int a,int b,int c,int d)
{
	if(c<a || d<b)
	{
		return 0;
	}
	else if(c==a && b==d)
	{
		return dp[c][d]=1;
	}
	else if(dp[c][d])
	{
		return dp[c][d];
	}
	
	dp[c][d]=(sana(a,b,c-1,d)*d+sana(a,b,c,d-1)*c-sana(a,b,c-1,d-1)*(c-1)*(d-1))%iris;
	dp[c][d]=(dp[c][d]+iris)%iris;
	return dp[c][d];
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	int a,b,c,d;
	cin>>a>>b>>c>>d;
	cout<<sana(a,b,c,d)<<'\n';
	
	return 0;
}