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<cstdio>
#include<iostream>
#include<cctype>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int N = 4e3 + 3;
const LL mod = 924844033; 
LL dp[N][N][2];
LL f[N];
bool check[N];
int n,k,cnt;
LL fac[N];
inline void up(LL &x){
	if(x >= mod) x -= mod;	
}
int main(){
	cin >> n >> k;
	for(int i = 1;i <= k;++i){
		for(int m = 1;m <= 2;++m){
			for(int j = i;j <= n;j += k){
				cnt++;
				if(j != i) check[cnt] = 1;	
			}
		}
	}
	dp[0][0][0] = 1;
	for(int i = 0;i < n * 2;++i)
		for(int j = 0;j <= n;++j){
			dp[i + 1][j][0] = dp[i][j][0] + dp[i][j][1];
			up(dp[i + 1][j][0]);
			if(check[i + 1]) dp[i + 1][j + 1][1] = dp[i][j][0];
		}
	fac[0] = 1;
	for(int i = 1;i <= n;++i)
		fac[i] = (fac[i - 1] * i) % mod;
	for(int i = 0;i <= n;++i)
		f[i] = (dp[2 * n][i][0] + dp[2 * n][i][1]) % mod;
	//for(int i = 0;i <= n;++i) printf("%lld ",fac[i]);puts("");
	LL ans = 0;	
	for(int i = 0;i <= n;++i){
		if(i & 1) ans = (ans - (f[i] * fac[n - i]) % mod + mod) % mod;
		else ans = (ans + f[i] * fac[n - i] % mod) % mod;
	}
	printf("%lld\n",ans);
	return 0;	
}