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;
using ll = long long;

const int mod = 998244353;
ll dp[5999][3000];

int a, b, c, d;
int main() {
	cin >> a >> b >> c >> d;
	dp[0][0] = 1;
	for (int k = 1; k <= c - a + d - b; ++k)
		for (int i = 0; i <= min(k, c - a); ++i) {
			int j = k - i;
			if (i > 0)
				dp[k][i] += dp[k - 1][i - 1] * (b + j);
			if (j > 0)
				dp[k][i] += dp[k - 1][i] * (a + i);
			if (i * j > 0)
				dp[k][i] -= dp[k - 2][i - 1] * (a + i - 1) * (b + j - 1);

			dp[k][i] %= mod;
		}
	if (dp[c - a + d - b][c - a] < 0)
		dp[c - a + d - b][c - a] += mod;
	cout << dp[c - a + d - b][c - a] << endl;
}