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 Int int64_t

using namespace std;

int main() {
	constexpr Int MOD = 998244353;
	Int A, B, C, D;
	cin >> A >> B >> C >> D;

	vector<vector<Int>> dp(C + 10, vector<Int>(D + 10, 0));
	dp[A][B] = 1;
	for (Int a = A; a <= C; ++a) {
		for (Int b = B; b <= D; ++b) {
			(dp[a + 1][b] += b * dp[a][b]) %= MOD;
			(dp[a][b + 1] += a * dp[a][b]) %= MOD;
			
			Int X = dp[a][b] * a % MOD * b % MOD;
			(dp[a + 1][b + 1] += MOD - X) %= MOD;
		}
	}
	cout << dp[C][D] % MOD << endl;

	return 0;
}