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>
#include<algorithm>
using namespace std;
#define MAXN 2010
#define MO 924844033
#define LL long long
int d[MAXN*2][MAXN][2],f[MAXN],n,k,tot,fac[MAXN],ans;
bool vis[MAXN][2],ed[MAXN*2];
int main()
{
	scanf("%d%d",&n,&k);
	fac[0]=1;
	for(int i=1;i<=n;i++)
		fac[i]=1LL*i*fac[i-1]%MO;
	for(int i=1;i<=k;i++)
	{
		tot+=(n-i)/k+1;
		ed[tot]=1;
		tot+=(n-i)/k+1;
		ed[tot]=1;
	}
	ed[0]=1;
	d[0][0][0]=1;
	for(int i=0;i<n*2;i++)
		for(int j=0;j<=n;j++)
		{
			d[i+1][j][0]=(d[i][j][0]+d[i][j][1])%MO;
			if(!ed[i])
				d[i+1][j+1][1]=d[i][j][0];
		}
	for(int i=0;i<=n;i++)
		f[i]=1LL*(d[n*2][i][0]+d[n*2][i][1])%MO*fac[n-i]%MO;
	for(int i=0,j=1;i<=n;j=-j,i++)
		(ans+=f[i]*j)%=MO;
	printf("%d\n",(ans%MO+MO)%MO);
}