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<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<bitset>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
const int N=3010,mod=998244353;
int A,B,C,D;
ll f[N][N],sum;
ll ksm(ll x,ll k){ll re=1;for(;k;k>>=1,x=x*x%mod)if(k&1)re=re*x%mod;return re;}
int main(){
	//freopen("a.in","r",stdin);
	scanf("%d%d%d%d",&A,&C,&B,&D);
	f[A][C]=1;
	for(int i=A+1;i<=B;i++){sum=0;
		for(int j=C;j<=D;j++){
			f[i][j]=(f[i-1][j]*j+sum*(i-1))%mod;
			sum=(sum*(i-1)+f[i-1][j])%mod;
		}
	}sum=0;
	for(int i=C;i<=D;i++)sum=(sum+f[B][i]*ksm(B,D-i))%mod;
	cout<<sum;
}