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>
typedef long long ll;

const int N=3005,P=998244353;
int A,B,C,D;
int f[N][N];

inline void Inc(int &x){x>=P?x-=P:0;}
ll Pow(ll a,ll b,ll s=1){for(;b;b>>=1,a=a*a%P)if(b&1)s=s*a%P;return s;}

int main()
{
	scanf("%d%d%d%d",&A,&B,&C,&D);
	for(int i=1;i<=3000;++i)
		for(int j=1;j<=3000;++j)
			if(i<A||j<B)continue;
			else if(i==A&&j==B)f[i][j]=1;
			else
				f[i][j]=((ll)f[i-1][j]*j+(ll)f[i][j-1]*i-(ll)f[i-1][j-1]*((i-1)*(j-1))%P+P)%P;
	/*for(int i=1;i<=C;++i)
		for(int j=1;j<=D;++j)
			printf("%d%c",f[i][j],j==D?'\n':' ');*/

	return printf("%d\n",f[C][D]),0;
}