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;

ll A, B, C, D;
vector<vector<vector<ll>>> Dp;
const ll MOD = 998244353;

int main() {
	cin >> A >> B >> C >> D;
	vector<vector<long long>> DP(C + 1, vector<long long >(D + 1));
	DP[A][B] = 1;
	for (int n = A; n <= C; ++n) {
		for (int m = B; m <= D; ++m) {
			if (n == A && m == B) {
				continue;
			}
			DP[n][m] = ((m) % MOD*DP[n - 1][m] % MOD + (n) % MOD*DP[n][m - 1] % MOD - (n - 1) % MOD*(m - 1) % MOD*DP[n - 1][m - 1] % MOD + MOD) % MOD;
		}
	}
	cout << DP[C][D] << endl;
	return 0;
}