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<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
using namespace std;
#define mod 924844033
int a[4001];
int prod[100001];
int f[4001][2001][2];
signed main()
{
	int n,k,cnt=0;
	scanf("%d%d",&n,&k);
	for(int i=1;i<=k;i++)
		for(int j=i;j<=n;j+=k)
			a[++cnt]=j;
	for(int i=1;i<=k;i++)
		for(int j=i;j<=n;j+=k)
			a[++cnt]=j;
	prod[0]=1;
	for(int i=1;i<=n;i++)
		prod[i]=1ll*prod[i-1]*i%mod;
	f[1][0][0]=1;
	for(int i=2;i<=2*n;i++)
	{
		for(int j = 0; j <= min(n, i/2); ++j)
		{
			f[i][j][0]=(f[i-1][j][1]+f[i-1][j][0])%mod;
			if(j>0&&a[i]==a[i-1]+k) f[i][j][1]=f[i-1][j-1][0];
		}
	}
	int ans=prod[n];
	for(int i=1;i<=n;i++)
	{
		int sum=(f[2*n][i][0]+f[2*n][i][1])%mod;
		ans+=1ll*((i%2==1)?-1:1)*sum%mod*prod[n-i]%mod;
		ans%=mod;
	}
	printf("%d",(ans+mod)%mod);
	return 0;
}