#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
typedef vector<vll> vvll;
#define forl(x,s,e) for(int x=s;x<e;x++)
#define all(x) x.begin(),x.end()
#define print(x) cout<<x<<'\n'
#define printv(x) for(auto c: x)cout<<c<<' ';cout<<'\n';

ll mod = 924844033;
ll MOD = 924844033;
ll N = 2101;
vll fact(N,1);

vll preprocess(ll n,ll k){

    ll kk = k*2;
    vvll dp;

    forl(r,0,kk){
        ll cnt = 0  ;                
        forl(j,0,n)
            if(j%kk==r)
                cnt+=1;
        if(cnt==0) break;
        vector<vvll> loc_dp(cnt+1,vvll(n+1,vll(3,0ll)));
        
        forl(ind,0,cnt+1)
            loc_dp[ind][0][0] = 1;
        if(cnt >=1){
            loc_dp[1][0][0] = 1;            
            loc_dp[1][1][1] = r-k >=0  ? 1:0;
            loc_dp[1][1][2] = r+k < n  ? 1:0;
        }

        forl(ind,2,cnt+1)
            forl(m,1,n+1){                
                ll index = r+(ind-1)*kk;
                loc_dp[ind][m][0] = (loc_dp[ind-1][m][0] + loc_dp[ind-1][m][1] + loc_dp[ind-1][m][2])%MOD;
                if(index+k<n)
                    loc_dp[ind][m][2] = (loc_dp[ind-1][m-1][0] + loc_dp[ind-1][m-1][1] + loc_dp[ind-1][m-1][2])%MOD;
                if(index-k>=0)
                    loc_dp[ind][m][1] = (loc_dp[ind-1][m-1][0] + loc_dp[ind-1][m-1][1])%MOD;                
            }

        vll val;
        forl(m,0,n+1)
            val.push_back(loc_dp[cnt][m][0]+loc_dp[cnt][m][1]+loc_dp[cnt][m][2]);
        dp.push_back(val)   ;     
    }
    ll sz = dp.size();
    vvll fdp(sz+1,vll(n+1,0ll));
    fdp[0][0] = 1;
    
    // Time complexity of the next few lines = O(sz*n*n//sz) the break statement will activate after part = n//sz 
    // as each residue group will have roughly n//sz elements in it
    forl(ind,1,sz+1)
        forl(m,0,n+1)
            forl(part,0,m+1){
                ll val1 = (dp[ind-1][part]*fdp[ind-1][m-part])%MOD;
                if(dp[ind-1][part]==0)break;// this break statement is crucial to prevent TLE
                fdp[ind][m] = (fdp[ind][m] + val1 )%MOD;
            }
    return fdp[sz];  
}


int main() {
    forl(i,2,N)
        fact[i] = (fact[i-1]*i)%mod;
    ll n,k;
    cin>>n>>k;

    vll dp = preprocess(n,k);
    ll ans = fact[n]; 
    forl(i,1,n+1){
        ll val = (dp[i]*fact[n-i])%mod;
        if(i%2==0)
            ans = (ans + val)%mod;                 
        else
            ans = (ans + mod - val)%mod; 
    }
    print(ans);
    return 0;
}