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<iostream>
#include<cstdio>
#define LL long long
using namespace std;

LL mod=998244353;
LL a,b,c,d;
int dp1[3005][3005],dp2[3005][3005];
int main()
{
	LL i,j;
	cin>>a>>b>>c>>d;
	dp1[a][b]=1;
	for(i=a+1;i<=c;i++)
		dp1[i][b]=b*dp1[i-1][b]%mod;
	for(i=b+1;i<=d;i++)
		dp1[a][i]=a*dp1[a][i-1]%mod;
	for(i=a+1;i<=c;i++)
		for(j=b+1;j<=d;j++)
		{
			dp1[i][j]=(dp1[i][j-1]+dp2[i][j-1])*i%mod;
			dp2[i][j]=(dp1[i-1][j]+dp2[i-1][j]*j)%mod;
		}
	cout<<(dp1[c][d]+dp2[c][d])%mod;
	return 0;
}