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:
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 3005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

ll dp[maxn][maxn],a,b,c,d;
const ll mod=998244353;

int main()
{
	cin>>a>>b>>c>>d; dp[a][b]=1;
	for(ll i=a;i<=c;i++)
	{
		for(ll j=b;j<=d;j++)
		{
			if(i==a&&j==b) continue;
			ll tmp=((i-1)*(j-1)*(dp[i-1][j-1]))%mod;
			dp[i][j]=((dp[i-1][j]*j)%mod+(i*dp[i][j-1])%mod-tmp+mod)%mod;
		}
	}
	cout<<dp[c][d]<<endl;
	return 0;
}