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;
typedef long long LL;
int A, B, C, D;
const int N = 6050;
const LL MOD = 998244353;
LL dp[N][N / 2];
void add(LL &x, LL y) {
	(x += y) %= MOD;
}
int main() {
	cin >> A >> B >> C >> D;
	dp[C][D] = 1;
	for (int i = C; i >= A; i --) {
		for (int j = D; j >= B; j --) {
			if (i == C && j == D) continue;
			dp[i][j] = dp[i+1][j] * j % MOD + dp[i][j+1] * i % MOD - i * j % MOD * dp[i+1][j+1] % MOD;
			dp[i][j] = (dp[i][j] % MOD + MOD) % MOD;
		}
	}
	cout << dp[A][B] << endl;
}