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>
#define IL inline
#define RG register
#define LL long long
#define Upd(a) (a)>=mod?(a)-=mod:1
using namespace std;
int n, k;
const int _ =2001, mod = 924844033;
bool Use[_<<1];
LL dp[_<<1][_][2], pre=0, pg=0;
LL jc[_], g[_];
IL int Add (RG int now) {
	if (now<=n) {
		if (now+k>n) return -1;
		return now+k+n;
	}
	else {
		if (now+k-n>n) return -1;
		return now+k-n;
	}
}
IL void work (RG int now) {
	for (RG int i=0; i<=n; ++i) dp[pre+1][i][0]=dp[pre][i][0]+dp[pre][i][1], Upd(dp[pre+1][i][0]);
	++pre;
	Use[now]=1;
	for (RG int i=Add(now); ~i; i=Add(i)) {
		Use[i]=1;
		for (RG int j=1; j<=n; ++j) {
			dp[pre+1][j][1]=dp[pre][j-1][0];
		}
		for (RG int j=0; j<=n; ++j) {
			dp[pre+1][j][0]=dp[pre][j][1]+dp[pre][j][0];
			Upd(dp[pre+1][j][0]);
		}
		++pre;
	}
}
int main () {
	cin>>n>>k;
	dp[0][0][0]=1;
	for (RG int i=1; i<=n; ++i) {
		if (!Use[i]) work (i);
		if (!Use[i+n]) work (i+n);
	}
	jc[0]=1;
	for (RG int i=1; i<=n; ++i) jc[i]=1LL*jc[i-1]*i%mod;
	for (RG int i=0; i<=n; ++i) g[i]=1LL*(dp[n*2][i][0]+dp[n*2][i][1])%mod*jc[n-i]%mod;
	for (RG int i=1, j=1; i<=n; ++i, j=-j) g[0]-=1LL*g[i]*j%mod, g[0]=(g[0]+mod)%mod;
	cout<<g[0];
}