#pragma GCC optimization ("O3")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;

#define INF (1LL<<61)
#define MOD 1000000007LL 
//#define MOD 998244353LL
#define EPS (1e-10)

#define PR(x) cout << (x) << endl
#define PS(x) cout << (x) << " "
#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))
#define FORE(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((ll)(x).size())
#define REV(x) reverse(ALL((x)))
#define ASC(x) sort(ALL((x)))
#define DESC(x) {ASC((x)); REV((x));}
#define BIT(s,i) (((s)>>(i))&1)
#define pb push_back
#define fi first
#define se second

template<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}
class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=MOD) x-=MOD; return *this;}
    mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}
    mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;}
    mint operator+(const mint& a) const {mint b(*this); return b+=a;}
    mint operator-(const mint& a) const {mint b(*this); return b-=a;}
    mint operator*(const mint& a) const {mint b(*this); return b*=a;}
    mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;}
    mint inv() const {return pow(MOD-2);}
    mint& operator/=(const mint& a) {return *this*=a.inv();}
    mint operator/(const mint& a) const {mint b(*this); return b/=a;}
};
istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;}
ostream &operator<<(ostream& os, const mint& a) {return os<<a.x;}
using mvec = vector<mint>;
using mmat = vector<mvec>;
using dvec = vector<double>;
using dmat = vector<dvec>;

ll modpow(ll a, ll n, ll m)
{
    if(n == 0) return 1;
    ll t = modpow(a, n>>1, m);
    return (n&1?t*a%m:t)*t%m;
}

int main()
{
    ll N, K;
    cin >> N >> K;

    vec ans(N);
    ll L = 0;
    while((1LL<<L) < 2*N+1) ++L;
    ll r = (modpow(2, K, N*2)+N*2-1)%(N*2);
    if(K >= L) {
        ll c = modpow(N*2, MOD-2, MOD)*(modpow(2, K, MOD)-1-r+MOD)%MOD;
        REP(i,0,N*2) {
            ll j = (i*2+N-modpow(2, K, N)+1)%N;
            ans[j] += c;
        }
    }
    REP(i,0,r+1) {
        ll j = (i*2+N-modpow(2, K, N)+1)%N;
        ++ans[j];
    }
    REP(i,0,N) PR(ans[i]*modpow(modpow(2,MOD-2,MOD),K,MOD)%MOD);

    return 0;  
}
/*



*/