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<cstdlib>
#include<cstring>
#include<iostream>
#define LL long long
using namespace std;
const LL mod=924844033;
LL n,k;
LL f[4010][2010][2];
LL vis[2010],a[4010],N=0;
void add(LL x,LL tim)
{
	LL i=x;a[N+1]=1;
	//printf("N:%d\n",N);
	while(1)
	{
		tim^=1;if(tim) vis[i]=1;N++;
		if(i+k>n) break;i+=k;
	}
}
LL fac[2010];
int main()
{
	scanf("%lld %lld",&n,&k);
	memset(vis,0,sizeof(vis));
	for(LL i=1;i<=n;i++) if(vis[i]==0) add(i,0),add(i,1);
	for(LL i=0;i<=N;i++) f[i][0][0]=1;
	for(LL i=1;i<=N;i++)
		for(LL j=1;j<=n;j++)
		{
			f[i][j][0]=(f[i-1][j][0]+f[i-1][j][1])%mod;
			if(a[i]!=1) f[i][j][1]=f[i-1][j-1][0];
		}
	//for(int i=1;i<=N;i++) printf("%lld ",a[i]);printf("\n");
	//for(LL i=0;i<=n;i++) printf("%lld:%lld\n",i,f[N][i][0]+f[N][i][1]);
	LL ans=0;
	fac[0]=1;for(LL i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
	for(LL i=0;i<=n;i++)
		(ans+=(i&1?-1:1)*fac[n-i]%mod*(f[N][i][0]+f[N][i][1])%mod)%=mod;
	printf("%lld",(ans+mod)%mod);
}