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<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;

const int N=4005;
const int mod=924844033;

int n,k,x,y,len,m,tag,ans;
int a[N],dp[N][N][2],f[N],c[N][N],fac[N];
bool vis[N][2];

void init(int n){
	int i,j;
	c[0][0]=1; fac[0]=1;
	for (i=1;i<=n;i++){
		c[i][0]=1;
		for (j=1;j<=i;j++) c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
	}
	for (i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mod;
}

int main(){
	int i,j;
	scanf("%d%d",&n,&k);
	init(4000);
	for (i=1;i<=n;i++)
		for (j=0;j<2;j++) if (!vis[i][j]){
			for (x=i,y=j;x<=n;x+=k,y^=1){
				vis[x][y]=1;
				m++;
			}
			a[m]=1;
		}
	dp[1][0][0]=1;
	for (i=1;i<m;i++)
		for (j=0;j<=i;j++){
			(dp[i+1][j][0]+=(dp[i][j][0]+dp[i][j][1])%mod)%=mod;
			if (!a[i]) (dp[i+1][j+1][1]+=dp[i][j][0])%=mod;
		}
	for (i=0;i<=m;i++) (f[i]+=dp[m][i][0]+dp[m][i][1])%=mod;
	ans=fac[n]; tag=-1; m=min(n,m);
	for (i=1;i<=m;i++){
		(ans+=1ll*tag*f[i]*fac[n-i]%mod)%=mod;
		tag*=-1;
	}
	ans=(ans+mod)%mod;
	printf("%d\n",ans);
	return 0;
}