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>
#define FOR(i,a,b) for(int i=a;i<=b;++i)

const int N=2005;
const int P=924844033;
int n,k,a[N],fac[N];
long long f[N][N][2][2];
bool vis[N];

int main() {
    scanf("%d%d",&n,&k);
    fac[0]=1;
    FOR(i,1,n) fac[i]=1LL*fac[i-1]*i%P;
    int tot=0;
    FOR(i,1,n) if(!vis[i]) for(int j=i;j<=n;j+=k) vis[j]=1,a[++tot]=j;
    f[0][0][0][0]=1;
    a[0]=-(1<<30);
    FOR(i,1,n) {
        f[i][0][0][0]=1;
        FOR(j,1,i) {
            f[i][j][0][0]=(f[i-1][j][1][0]+f[i-1][j][0][0]+(a[i]-a[i-1]==k)*f[i-1][j-1][0][0])%P; // i丢给i-1
            f[i][j][0][1]=(a[i+1]-a[i]==k)*(f[i-1][j-1][1][0]+f[i-1][j-1][0][0])%P; // i+1丢给i
            f[i][j][1][0]=(f[i-1][j][0][1]+f[i-1][j][1][1]+(a[i]-a[i-1]==k)*f[i-1][j-1][0][1])%P; // i-1丢给i
            f[i][j][1][1]=(a[i+1]-a[i]==k)*(f[i-1][j-1][0][1]+f[i-1][j-1][1][1])%P;
        }
    }
    int ans=0;
    FOR(i,0,n) {
        int sum=(f[n][i][0][0]+f[n][i][0][1]+f[n][i][1][0]+f[n][i][1][1])%P;
        if(i&1) (ans+=P-1LL*sum*fac[n-i]%P)%=P;
        else (ans+=1LL*sum*fac[n-i]%P)%=P;
    }
    printf("%d\n",ans);
    return 0;
}