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;

int n,k,a[4010],cnt;
long long ans,fac[2010],dp[4010][2010][2];
const long long mod=924844033;
int main()
{
	scanf("%d%d",&n,&k);
	fac[0]=1;
	for(int i=1;i<=n;++i)
	fac[i]=fac[i-1]*i%mod;
	for(int i=1;i<=k;++i)
	{
		for(int l=1;l<=2;++l)//有对称的两条链，举个例子，一个是1->3'->5，一个是1'->3->5' 
		{
			for(int j=i;j<=n;j+=k)
			{
				++cnt;
				if(j!=i)//链头不与前面一个连 
				a[cnt]=1;
			}
		}			
	}
	dp[0][0][0]=1;
	for(int i=0;i<2*n;++i)
	{
		for(int j=0;j<=n;++j)
		{
			dp[i+1][j][0]=(dp[i][j][0]+dp[i][j][1])%mod;
			if(a[i+1])
			dp[i+1][j+1][1]=dp[i][j][0];
		}
	}
	for(int i=0;i<=n;++i)
	{
		if(i&1)
		ans=(ans-(dp[2*n][i][0]+dp[2*n][i][1])*fac[n-i]%mod+mod)%mod;
		else
		ans=(ans+(dp[2*n][i][0]+dp[2*n][i][1])*fac[n-i])%mod;
	}
	printf("%lld\n",ans);
	return 0;
}