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:
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
using namespace std;
const int P=924844033;
int n,k,ed[4010],dp[4010][2010][2],fac[4010],ans,cnt;
int main(){
	scanf("%d%d",&n,&k);
	rep(i,1,k) rep(j,0,1){
		for(int w=i,t=j;w<=n;w+=k,t^=1) ++cnt;
		ed[cnt]=1;
	}dp[1][0][0]=1;
	rep(i,1,n*2) rep(j,0,n){
		dp[i+1][j][0]=(dp[i][j][0]+dp[i][j][1])%P;
		if(!ed[i]) dp[i+1][j+1][1]=dp[i][j][0];
	}
	fac[0]=1;rep(i,1,n) fac[i]=1ll*fac[i-1]*i%P;
	rep(i,0,n)
		if(i&1) ans=(ans-1ll*(dp[n*2][i][0]+dp[n*2][i][1])%P*fac[n-i]%P+P)%P;
		else ans=(ans+1ll*(dp[n*2][i][0]+dp[n*2][i][1])%P*fac[n-i]%P)%P;
	printf("%d\n",ans);
	return 0;
}