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;

constexpr long long MOD = 998244353;

int main() {
	long long A, B, C, D;
	cin >> A >> B >> C >> D;
	static long long DP[3001][3001] = {};
	DP[A][B] = 1;
	for (int i = A; i < C; i++) {
		DP[i + 1][B] = (DP[i][B] * B) % MOD;
	}
	for (int i = B; i < D; i++) {
		DP[A][i + 1] = (DP[A][i] * A) % MOD;
	}
	for (int i = A + 1; i <= C; i++) {
		for (int j = B + 1; j <= D; j++) {
			DP[i][j] = ((DP[i - 1][j] * j + DP[i][j - 1] * i) % MOD - (DP[i - 1][j - 1] * (i - 1) * (j - 1)) % MOD + MOD) % MOD;
		}
	}
	cout << DP[C][D] << endl;
}