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>
#define ll long long
using namespace std;
const int maxn=3e3+10;
const ll mod=998244353;
ll dp[maxn][maxn];

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