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>
#define mod 924844033
typedef long long LL;
const int N=4005;//2n

int ban[N],f[2][N][2],fac[N];

int main()
{
	int n,K,cnt=0; scanf("%d%d",&n,&K);
	for(int i=1; i<=K; ++i)
	{
		ban[++cnt]=1;
		for(int j=i+K; j<=n; j+=K) ban[++cnt]=0;
		ban[++cnt]=1;//就是两条链啊 
		for(int j=i+K; j<=n; j+=K) ban[++cnt]=0;
	}
	int p=0; f[p][0][0]=1;
	for(int i=0; i<cnt; ++i,p^=1)
		for(int j=0; j<=i; ++j)
		{
			f[p^1][j][0]=(f[p][j][0]+f[p][j][1])%mod;
			if(!ban[i+1]) f[p^1][j+1][1]=f[p][j][0];
			f[p][j][0]=f[p][j][1]=0;
		}
	fac[0]=1;
	for(int i=1; i<=n; ++i) fac[i]=1ll*fac[i-1]*i%mod;
	LL ans=0;
	for(int i=0; i<=n; ++i)
		ans+=i&1?mod-1ll*fac[n-i]*(f[p][i][0]+f[p][i][1])%mod:1ll*fac[n-i]*(f[p][i][0]+f[p][i][1])%mod;
	printf("%lld\n",ans%mod);

	return 0;
}