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
const int maxn=3010,mod=998244353;
int a,b,c,d,dp[maxn][maxn];
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>a>>b>>c>>d;
	dp[a][b]=1;
	for(int i=a;i<=c;i++)
		for(int j=b;j<=d;j++)
			if(i!=a||j!=b)
			{
				if(i==1||j==1)
					dp[i][j]=1;
				else
					dp[i][j]=(dp[i-1][j]*j%mod+dp[i][j-1]*i%mod-dp[i-1][j-1]*(i-1)%mod*(j-1)%mod+mod)%mod;
			}
	cout<<dp[c][d]<<endl;
	return 0;
}