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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2005,P=924844033,M=4005;
int n,k,f[M][N],l[N],a[N],col[M],cnt,pos,fac[N],ans;
int main(){
	scanf("%d%d",&n,&k);fac[0]=1;
	for(int i=1;i<=n;i++)l[i%k]++,fac[i]=(ll)fac[i-1]*i%P;
	for(int i=0;i<k;i++)a[l[i]]+=2;
	for(int i=1;i<=n;i++)while(a[i]){
		++cnt;for(int j=1;j<=i;j++)col[++pos]=cnt;a[i]--;
	}
	f[0][0]=1;
	for(int i=1;i<=n*2;i++){
		int*a=f[i],*b=f[i-1];memcpy(a,b,(n+1)<<2);
		if(col[i]==col[i-1]){int*c=f[i-2];for(int j=1;j<=n;j++)(a[j]+=c[j-1])%=P;}
	}
	int*num=f[n*2];
	for(int i=0;i<=n;i++)ans=(ans+(ll)num[i]*fac[n-i]*((i&1)?-1:1))%P;
	printf("%d\n",(ans+P)%P);
	return 0;
}