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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=3005;
const int MOD=998244353;
int A,B,C,D;
long long dp[N][N];
long long dfs(int r,int c)
{
	if(r<A||c<B) return 0;
	if(dp[r][c]!=-1) return dp[r][c];
	int res=0;
	res=(res+dfs(r,c-1)*r%MOD)%MOD;
	res=(res+dfs(r-1,c)*c%MOD)%MOD;
	res=(res-dfs(r-1,c-1)*(r-1)%MOD*(c-1)%MOD+MOD)%MOD;
	return dp[r][c]=res;
}
int main()
{
	scanf("%d%d%d%d",&A,&B,&C,&D);
	memset(dp,-1,sizeof(dp));
	dp[A][B]=1;
	printf("%lld",dfs(C,D));
	return 0;
}