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<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<complex>
#include<iostream>
#include<algorithm>
#define LL long long
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
	while(ch=getchar(),ch<33);
	if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
	while(ch=getchar(),ch>33)x=x*10+ch-48;
	if(fs)x=-x;	
}
using namespace std;
const int mod=924844033;
int n,k,ans,fac[2001],g[2001],f[4001][2001][2];
int main()
{
	re(n),re(k);
	f[0][0][0]=1;
	for(int i=1,be=2,now=1;i<=n+n;++i)
	{
		for(int j=0;j<=n;++j)	
		{
			f[i][j][0]=(f[i-1][j][1]+f[i-1][j][0])%mod;
			if(now!=be/2)f[i][j][1]=f[i-1][j-1][0];
		}
		now+=k;
		if(now>n)
			++be,now=be/2;
	}
	for(int i=0;i<=n;++i)g[i]=(f[n+n][i][0]+f[n+n][i][1])%mod;
	fac[0]=1;
	for(int i=1;i<=n;++i)
		fac[i]=1ll*fac[i-1]*i%mod;
	for(int i=0,f=1;i<=n;++i,f*=-1)	
		ans=(ans+1ll*f*fac[n-i]*g[i])%mod;
	ans=(ans+mod)%mod;
	printf("%d\n",ans);
}