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: Accepted)

#include <bits/stdc++.h>
using namespace std;

using ll=long long;
const ll MOD=924844033;
ll ans;
ll dp[2234][2234];//dp[i][j]:i-1本目まで見てj本選ぶ
ll DP[5000][5000];//DP[i]:i本選んだ通り
ll F[5000];

signed main(){
  int n,k;cin>>n>>k;

  F[0]=1;
  for(int i=1;i<5000;i++)F[i]=F[i-1]*i%MOD;

  dp[0][0]=dp[1][0]=dp[1][1]=1;
  for(int i=1;i<2200;i++){
    for(int j=0;j<2200;j++){
      (dp[i+1][j]+=dp[i][j])%=MOD;
      (dp[i+1][j+1]+=dp[i-1][j])%=MOD;
    }
  }
  DP[0][0]=1;
  for(int i=0;i<k;i++){
    int U=(n-i-k-1)/(k*2)+1,D=(n-i-1)/(k*2)+1;
    if(n-i-k-1<0)U=0;
    if(n-i-1<0)D=0;
    int A=U+D-1;
    for(int j=2000;j>=0;j--){
      for(int l=0;l<=A;l++){
        (DP[i+1][j+l]+=DP[i][j]*dp[A][l])%=MOD;
      }
    }
  }
  for(int i=0;i<k;i++){
    int U=(n-i-k-1)/(k*2)+1,D=(n-i-1)/(k*2)+1;
    if(n-i-k-1<0)U=0;
    if(n-i-1<0)D=0;
    int A=U+D-1;
    for(int j=2000;j>=0;j--){
      for(int l=0;l<=A;l++){
        (DP[i+k+1][j+l]+=DP[i+k][j]*dp[A][l])%=MOD;
      }
    }
  }
  ll ans=0;
  for(int i=0;i<=n;i++){
    ll tmp=DP[2*k][i]*F[n-i]%MOD;
    if(i&1)tmp*=-1;
    tmp+=MOD;
    (ans+=tmp)%=MOD;
  }
  cout<<ans<<endl;
}