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 <stdio.h>
long long p = 998244353;
long long a, b, c, d;
long long x[3001][3001];
int main() {
	scanf("%lld%lld%lld%lld", &a, &b, &c, &d);
	x[a][b] = 1;
	for (long long i = a + 1; i <= c; i++) {
		x[i][b] = x[i - 1][b] * b % p;
	}
	for (long long i = b + 1; i <= d; i++) {
		x[a][i] = x[a][i - 1] * a % p;
	}
	for (long long i = a + 1; i <= c; i++) {
		for (long long j = b + 1; j <= d; j++) {
			x[i][j] = (x[i - 1][j] * j + x[i][j - 1] * i) % p;
			x[i][j] -= x[i - 1][j - 1] * (i - 1) * (j - 1) % p;
			x[i][j] += p;
			x[i][j] %= p;
		}
	}
	printf("%lld\n", x[c][d]);
}