Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

3 1

Sample Output 1

2

2 permutations (1, 2, 3), (3, 2, 1) satisfy the condition.

Sample Input 2

4 1

Sample Output 2

5

5 permutations (1, 2, 3, 4), (1, 4, 3, 2), (3, 2, 1, 4), (3, 4, 1, 2), (4, 2, 3, 1) satisfy the condition.

Sample Input 3

4 2

Sample Output 3

9

Sample Input 4

4 3

Sample Output 4

14

Sample Input 5

425 48

Sample Output 5

756765083

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long ll;
const int MOD=924844033;
const int MAXN=2005;

int n,m,ncnt;
int ed[MAXN*2];
ll f[MAXN],d[MAXN][MAXN][2],fac[MAXN];
bool vis[MAXN][2];

void Init(){
    fac[0]=1;
    for(int i=1;i<MAXN;i++)
        fac[i]=fac[i-1]*i%MOD;
}

int main(){
    //freopen("captain.in","r",stdin);
    //freopen("captain.out","w",stdout);

    Init();
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=0;j<=1;j++){
            if(vis[i][j]) continue;
            for(int x=i,y=j;x<=n;x+=m,y^=1)
                vis[x][y]=1,ncnt++;
            ed[ncnt]=1;
        }
    ed[0]=1;
    d[0][0][0]=1;
    for(int i=0;i<2*n;i++)
        for(int j=0;j<=n;j++){
            d[i+1][j][0]=(d[i][j][0]+d[i][j][1])%MOD;
            if(!ed[i]) d[i+1][j+1][1]=d[i][j][0];
        }
    for(int i=0;i<=n;i++) f[i]=(d[2*n][i][0]+d[2*n][i][1])%MOD;
    ll ans=0;
    for(int i=0,k=1;i<=n;i++,k*=-1)
        ans=(ans+1ll*k*fac[n-i]*f[i])%MOD;
    printf("%lld\n",(ans+MOD)%MOD);

    fclose(stdin);
    fclose(stdout);
}