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;
const int mod = 998244353;
int A, B, C, D;
int main() {
	cin >> A >> B >> C >> D;
	vector<vector<ll>> b(C+1, vector<ll>(D+1));
	b[A][B] = 1;
	for (int i = A+1; i <= C; ++ i) {
		b[i][B] = b[i-1][B] * B % mod;
	}
	for (int i = B+1; i <= D; ++ i) {
		b[A][i] = b[A][i-1] * A % mod;
	}
	for (int i = A+1; i <= C; ++ i) {
		for (int j = B+1; j <= D; ++ j) {
			(b[i][j] += b[i-1][j] * j) %= mod;
			(b[i][j] += b[i][j-1] * i) %= mod;
			(b[i][j] += mod - b[i-1][j-1] * (i-1) % mod * (j-1) % mod) %= mod;
		}
	}
	cout << b[C][D] << endl;
}