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:
//2017-10-25
//miaomiao
//
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>

using namespace std;

#define For(i, a, b) for(int i = (a); i <= (int)(b); ++i)
#define N (4000+5)

const int P = 924844033;

int f[N][N][2], g[N], fac[N];
bool vis[N][2], ban[N];

int main(){
	int n, k;
	scanf("%d%d", &n, &k);

	int nn = 0;
	For(i, 1, n) For(j, 0, 1) if(!vis[i][j]){
		int x = i, y = j, len = 0;
		while(x <= n){
			vis[x][y] = true;
			++len; x += k; y ^= 1;
		}
		ban[nn+=len] = true;
	}

	f[1][0][0] = 1;
	For(i, 1, nn) For(j, 0, n){
		f[i+1][j][0] = (f[i][j][0] + f[i][j][1]) % P;
		if(!ban[i]) f[i+1][j+1][1] = f[i][j][0];
	}

	For(i, 0, n) g[i] = (f[nn][i][0] + f[nn][i][1]) % P;
	
	fac[0] = 1;
	For(i, 1, n) fac[i] = 1ll * fac[i-1] * i % P;

	int ans = fac[n];
	for(int i=1, j=-1; i <= n; ++i, j=-j) ans = ((ans + 1ll*j*g[i]*fac[n-i]%P)%P + P) % P;
	printf("%d\n", ans);

	return 0;
}