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<cstdio>
#define RI register int
#define CI const int&
using namespace std;
const int N=5005,mod=998244353;
int a,b,c,d,f[N][N];
int main()
{
	RI i,j; scanf("%d%d%d%d",&a,&b,&c,&d); f[a][b]=1;
	for (i=a;i<=c;++i) for (j=b;j<=d;++j)
	if (i>a&&j>b) f[i][j]=(1LL*f[i-1][j]*j%mod+1LL*f[i][j-1]*i%mod-1LL*(i-1)*(j-1)*f[i-1][j-1]%mod+mod)%mod;
	else if (i>a) f[i][j]=1LL*f[i-1][j]*j%mod; else if (j>b) f[i][j]=1LL*f[i][j-1]*i%mod;
	return printf("%d",f[c][d]),0;
}