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;

#define INF 1234567890
#define ll long long
#define MOD 998244353

int A, B, C, D;
ll dp[3030][3030];

int main()
{
	scanf("%d %d %d %d", &A, &B, &C, &D);
	for(int i=A; i<=C; i++)
		for(int j=B; j<=D; j++)
		{
			if (i == A && j == B)
			{
				dp[i][j] = 1;
				continue;
			}
			dp[i][j] = dp[i-1][j]*j + dp[i][j-1]*i - dp[i-1][j-1]*(i-1)*(j-1);
			dp[i][j] %= MOD;
		}
	ll res = dp[C][D];
	res += MOD; res %= MOD;
	printf("%lld\n", res);
	return 0;
}