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 md=998244353;
long long a,b,c,d,dp[2][3030][3030];
int main()
{
	cin>>a>>b>>c>>d;
	dp[0][a][b]=1;
	for(int i=a;i<=c;i++)
	{
		for(int j=b;j<=d;j++)
		{
			if(i==a && j==b)
			{
				continue;
			}
			dp[0][i][j]=dp[0][i-1][j]*j+dp[1][i-1][j];
			dp[0][i][j]%=md;
			dp[1][i][j]=dp[0][i][j-1]*i+dp[1][i][j-1]*i;
			dp[1][i][j]%=md;
		}
	}
	cout<<(dp[0][c][d]+dp[1][c][d])%md<<endl;
	return 0;
}