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<bits/stdc++.h>
using namespace std;
#define mod 924844033
#define N 2005
int n,k,cnt;
long long fac[N],f[N][N][2][2],num[N];
bool s[N];
int main()
{
	scanf("%d%d",&n,&k);fac[1]=1;fac[0]=1;
	for(int i=2;i<=n;i++) fac[i]=(fac[i-1]*1LL*i)%mod;
	for(int i=1;i<=n;i++)if(!s[i])for(int j=i;j<=n;j+=k) s[j]=1,num[++cnt]=j;
	f[0][0][0][0]=1;
	num[0]=-(1<<30);
	for(int i=1;i<=n;i++)//链 
	{
		f[i][0][0][0]=1;
		for(int j=1;j<=i;j++)
		{//继承，或后面放前面 
			f[i][j][0][0]=(f[i-1][j][1][0]+f[i-1][j][0][0]+((num[i]-num[i-1])==k)*f[i-1][j-1][0][0])%mod;
		 //继承 
			f[i][j][1][0]=(f[i-1][j][0][1]+f[i-1][j][1][1]+((num[i]-num[i-1])==k)*f[i-1][j-1][0][1])%mod;
		 //前面放后面 
			f[i][j][0][1]=((num[i+1]-num[i])==k)*(f[i-1][j-1][1][0]+f[i-1][j-1][0][0])%mod;
			f[i][j][1][1]=((num[i+1]-num[i])==k)*(f[i-1][j-1][0][1]+f[i-1][j-1][1][1])%mod; 
		}
	}
	long long ans=0;
	for(int i=0;i<=n;i++)
	{
		long long sum=(f[n][i][0][0]+f[n][i][0][1]+f[n][i][1][0]+f[n][i][1][1])%mod;
		if(i&1) ans=(ans+mod-(sum*fac[n-i])%mod)%mod;
		else ans=(ans+(sum*fac[n-i])%mod)%mod;
	}
	cout<<ans<<endl;
	return 0;
}