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 = 3005, mod = 998244353;

int n, m, dp[N][N];
int yo(int i, int j) {
	if (i < n || j < m) return 0;
	if (i == n && j == m) return 1;
	int &ret = dp[i][j];
	if (ret != -1) return ret;
	long long ans = 1LL * yo(i - 1, j) * j + 1LL * yo(i, j - 1) * i;
	ans -= 1LL * yo(i - 1, j - 1) * (i - 1) * (j - 1);
	ans %= mod;
	ans = (ans + mod) % mod;
	return ret = ans;
}
int32_t main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int p, q; cin >> n >> m >> p >> q;
	memset(dp, -1, sizeof dp);
	cout << yo(p, q) << '\n';
    return 0;
}