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;
#define int long long
#define mod 998244353
int dp[3005][3005];
 
int32_t main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	int a,b,c,d;
	cin>>a>>b>>c>>d;
	int n=c-a+1;
	int m=d-b+1;
	dp[1][1]=1;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			if(i==1&&j==1) continue;
			else
			{
				dp[i][j]=(dp[i-1][j]*(j+b-1)+dp[i][j-1]*(i+a-1)-dp[i-1][j-1]*(j-1+b-1)%mod*(i-1+a-1)%mod+mod)%mod;
			}
		}
	}
	cout<<dp[n][m]<<"\n";
	return 0;
}