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;
const int maxn=3010,mod=998244353;
inline void Add(int &a,int b){a=a+b>=mod?a+b-mod:a+b;}
inline int ksm(int a,int b)
{
	int res=1;
	for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) res=1ll*res*a%mod;
	return res;
}
int A,B,C,D,f[maxn][maxn];
int main()
{
	cin>>A>>B>>C>>D;
	f[A][B]=1;
	for(int i=A;i<=C;i++) for(int j=B;j<=D;j++)
	{
		Add(f[i+1][j],1ll*f[i][j]*j%mod);
		Add(f[i][j+1],1ll*f[i][j]*i%mod);
		Add(f[i+1][j+1],mod-1ll*f[i][j]*i%mod*j%mod);
	}
	cout<<f[C][D]<<endl;
}