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 <cstdio>
using namespace std;

typedef long long LL;

const int N = 2010;
const LL mod = 924844033;

int n, k;

bool vis[N][2], st[N << 1];
int cnt;

LL f[N << 1][N][2];
LL fac[N];

int main()
{
	scanf("%d %d", &n, &k);
	st[1] = 1;
	for (int i = 1; i <= n; ++ i)
	{
		for (int j = 0; j <= 1; ++ j)
		{
			if (vis[i][j]) continue;
			for (int x = i, y = j; x <= n; x += k, y ^= 1)
				vis[x][y] = 1, ++ cnt;
			st[cnt + 1] = 1;
		}
	}
	f[0][0][0] = 1;
	for (int i = 1; i <= n << 1; ++ i)
	{
		for (int j = 0; j <= i >> 1; ++ j)
		{
			f[i][j][0] = (f[i - 1][j][0] + f[i - 1][j][1]) % mod;
			if (!st[i] && j) f[i][j][1] = f[i - 1][j - 1][0];
		}
	}
	fac[0] = 1;
	for (int i = 1; i <= n; ++ i)
		fac[i] = fac[i - 1] * i % mod;
	LL ans = 0;
	for (int i = 0; i <= n; ++ i)
		(ans += (i & 1 ? - 1 : 1) * (f[n << 1][i][0] + f[n << 1][i][1]) % mod * fac[n - i] % mod + mod) %= mod;
	printf("%lld\n", ans);
}