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;
const int mod = 998244353;
int a, b, c, d;
int f[4000][4000][2];

void upd(int &a, int b) {
	a += b; if (a >= mod) a -= mod;
}

int main() {
	cin >> a >> b >> c >> d;
	f[a][b][0] = 1;
	for (int i = a; i <= c; ++i)
		for (int j = b; j <= d; ++j) {
			upd(f[i + 1][j][0], 1ll * f[i][j][0] * j % mod);
			upd(f[i][j + 1][1], 1ll * f[i][j][0] * i % mod);
			upd(f[i + 1][j][0], f[i][j][1]);
			upd(f[i][j + 1][1], 1ll * f[i][j][1] * i % mod);
		}
	printf("%d", (f[c][d][0] + f[c][d][1]) % mod);
}