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 Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
#define Reg register
#define RI Reg int
#define Con const
#define CI Con int&
#define I inline
#define W while
#define N 3000
#define X 998244353
using namespace std;
int a,b,c,d,f[N+5][N+5];
int main()
{
	RI i,j;scanf("%d%d%d%d",&a,&b,&c,&d);
	for(f[a][b]=1,i=a;i<=c;++i) for(j=b;j<=d;++j) if(a==i&&b==j);else
		if(b==j) f[i][j]=1LL*f[i-1][j]*j%X;else if(a==i) f[i][j]=1LL*f[i][j-1]*i%X;
		else f[i][j]=(1LL*f[i-1][j]*j+1LL*f[i][j-1]*i-1LL*(i-1)*(j-1)%X*f[i-1][j-1]%X+X)%X;
	return printf("%d\n",f[c][d]),0;
}