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>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=998244353;
inline int addmod(int x)
{
	return x>=mod?x-mod:x;
}
inline int submod(int x)
{
	return x<0?x+mod:x;
}
int a,b,c,d,f[3005][3005];
int main()
{
	scanf("%d%d%d%d",&a,&b,&c,&d);
	f[a][b]=1;
	for(int i=b+1;i<=d;i++)
		f[a][i]=1ll*f[a][i-1]*a%mod;
	for(int i=a+1;i<=c;i++)
		f[i][b]=1ll*f[i-1][b]*b%mod;
	for(int i=a+1;i<=c;i++)
		for(int j=b+1;j<=d;j++)
			f[i][j]=(1ll*f[i][j-1]*i%mod+1ll*f[i-1][j]*j%mod-1ll*f[i-1][j-1]*(i-1)*(j-1)%mod+mod)%mod;
	printf("%d\n",f[c][d]);
	return 0;		
}