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>
using namespace std;
typedef long long ll;
const int mod = 998244353;
int a,b,c,d;
ll mem[3005][3005];
ll dp(int x, int y){
	if (x == a && y == b) return 1ll;
	if (mem[x][y] != -1) return mem[x][y];
	ll res = 0;
	if (x > a){
		res += y*dp(x-1,y);
	}
	if (y > b){
		res += x*dp(x,y-1);
	}
	if (x > a && y > b) res -= (x-1)*(y-1)*dp(x-1,y-1);
	res = ((res % mod) + mod )%mod;
	return mem[x][y] = res;
}

int main(){
	scanf("%d%d%d%d",&a,&b,&c,&d);
	memset(mem,-1,sizeof(mem));
	printf("%lld",dp(c,d));
	
}