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;

const int N = 3e3 + 5;
typedef long long LL;

LL mod = 998244353, f[N][N];
int a, b, c, d;

LL cal(int x, int y) {
	return (x - 1)*(y - 1);
}

int main() {
	cin>>a>>b>>c>>d;

	f[a][b] = 1;
	for(int i = a; i <= c; i++)
		for(int j = b; j <= d; j++) 
			if(i + j != a + b) f[i][j] = (f[i - 1][j]*j%mod + f[i][j - 1]*i%mod - f[i - 1][j - 1]*cal(i, j)%mod + mod)%mod;

	cout<<f[c][d]%mod<<endl;
	return 0;
}