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 rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

const int MOD=998244353;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;

ll dp[5000][5000];

int main(){
	int a,b,c,d;cin>>a>>b>>c>>d;
	dp[a][b]=1;
	for(int i=a;i<c;i++){
		ll sum=0;
		for(int j=b;j<=d;j++){
			(dp[i+1][j]+=dp[i][j]*j)%=MOD;
			(dp[i+1][j]+=sum)%=MOD;
			(sum+=dp[i][j])%=MOD;
			(sum*=i)%=MOD;
		}
	}
	ll ans=0;
	ll D=1;
	for(int j=d;j>=b;j--){
		(ans+=dp[c][j]*D)%=MOD;
		(D*=c)%=MOD;
	}
	cout<<ans<<endl;
}