Problem: Snuke loves permutations. He is making a permutation of length N.

Since he hates the integer K, his permutation will satisfy the following:

* Let the permutation be a_1, a_2, ..., a_N. For each i = 1,2,...,N, |a_i - i| \neq K.



Among the N! permutations of length N, how many satisfies this condition?

Since the answer may be extremely large, find the answer modulo 924844033(prime).

Constraints

* 2 ≦ N ≦ 2000
* 1 ≦ K ≦ N-1

Input

The input is given from Standard Input in the following format:


N K


Output

Print the answer modulo 924844033.

Examples

Input

3 1


Output

2


Input

4 1


Output

5


Input

4 2


Output

9


Input

4 3


Output

14


Input

425 48


Output

756765083

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

const int Maxn = 2005, p = 924844033;
int n, k, ct, arr[Maxn];
long long ans, f[Maxn][Maxn][2][2], fac[Maxn];
int main()
{
	scanf("%d%d", &n, &k);
	fac[0] = 1;
	for (int i = 1; i <= n; i++)
		fac[i] = fac[i - 1] * i % p;
	for (int i = 1; i <= k; i++)
		for (int j = i; j <= n; j += k)
			arr[++ct] = j;
	arr[0] = -0x3f3f3f3f;
	f[0][0][0][0] = 1;
	for (int i = 1; i <= n; i++)
	{
		f[i][0][0][0] = 1; 
		for (int j = 1; j <= i; j++)
		{
			f[i][j][0][0] = (f[i - 1][j][0][0] + f[i - 1][j][1][0] + f[i - 1][j - 1][0][0] * (arr[i] - arr[i - 1] == k)) % p;
			f[i][j][0][1] = (arr[i + 1] - arr[i] == k) * (f[i - 1][j - 1][0][0] + f[i - 1][j - 1][1][0]) % p;
			f[i][j][1][0] = (f[i - 1][j][0][1] + f[i - 1][j][1][1] + (arr[i] - arr[i - 1] == k) * f[i - 1][j - 1][0][1]) % p;
			f[i][j][1][1] = (arr[i + 1] - arr[i] == k) * (f[i - 1][j - 1][0][1] + f[i - 1][j - 1][1][1]) % p;
		}
	}
	for (int i = 0; i <= n; i++)
		(ans += (i & 1 ? p - 1 : 1) * fac[n - i] % p * (f[n][i][0][0] + f[n][i][1][0])) %= p;
	printf("%lld", ans);
	return 0;
}