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 MOD 998244353LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int a,b,c,d;
ll dp[3005][3005];

int main(void){
	scanf("%d%d%d%d",&a,&b,&c,&d);
	dp[a][b]=1;
	for(int i=a;i<=c;i++){
		for(int j=b;j<=d;j++){
			dp[i][j]+=(ll)dp[i-1][j]*j%MOD;
			dp[i][j]+=(ll)dp[i][j-1]*i%MOD;
			ll val=dp[i-1][j-1]*(i-1)*(j-1)%MOD;
			dp[i][j]+=(MOD-val);
			dp[i][j]%=MOD;
		}
	}
	printf("%lld\n",dp[c][d]);
	return 0;
}