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 int long long
const int MOD = 998244353;
int dp[3001][3001];
void add(int &a, const int &b) {
	a += b;
	if (a >= MOD) a -= MOD;
}
void sub(int &a, const int &b) {
	a -= b;
	if (a < 0) a += MOD;
}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int A, B, C, D;
	cin >> A >> B >> C >> D;
	dp[A][B] = 1;
	for (int i = A; i <= C; i++) {
		for (int j = B; j <= D; j++) {
			add(dp[i][j], dp[i - 1][j] * j % MOD);
			add(dp[i][j], dp[i][j - 1] * i % MOD);
			sub(dp[i][j], dp[i - 1][j - 1] * (i - 1) * (j - 1) % MOD);
		}
	}
	cout << dp[C][D] << endl;
}