// Original C++ solution (slower):
// abc132_d

#include <bits/stdc++.h>

#ifdef LOCAL

#include "../cxx-prettyprint/prettyprint.hpp"

#endif

using namespace std;



typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> P;



#define REP(i, n) for (int (i) = 0 ; (i) < (int)(n) ; ++(i))

#define REPN(i, m, n) for (int (i) = m ; (i) < (int)(n) ; ++(i))

#define REP_REV(i, n) for (int (i) = (int)(n) - 1 ; (i) >= 0 ; --(i))

#define REPN_REV(i, m, n) for (int (i) = (int)(n) - 1 ; (i) >= m ; --(i))

#define ALL(x) x.begin(), x.end()



#define INF ((1 << 29)-1)

#define MOD (1000000007)



#define print2D(h, w, arr) REP(i, h) { REP(j, w) cout << arr[i][j] << " "; cout << endl; }

template<class T> void print(const T& x){cout << x << endl;}

template<class T, class... A> void print(const T& first, const A&... rest) { cout << first << " "; print(rest...); }

struct PreMain {PreMain(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(20);}} premain;





int mod_pow(ll a,ll b,int mo){int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}



ll factorial[200001];

void make_factorial(int mo){ factorial[0] = 1; REPN(i, 1, 200001) factorial[i] = (i * factorial[i-1]) % mo; }

ll get_factorial(int n){ return n >= 0 ? factorial[n] : 0; }



ll nCk(int n, int k, int mo){

    ll a = get_factorial(n);

    ll b = get_factorial(n-k) * get_factorial(k) % mo;

    ll b_inv = mod_pow(b, mo-2, mo);

    return a * b_inv % mo;

}



int main() {

#ifdef LOCAL

    ifstream in("../arg.txt"); cin.rdbuf(in.rdbuf());

#endif



    int N, K;

    cin >> N >> K;



    int b = K;

    int r = N - K;



    make_factorial(MOD);



    REPN(i, 1, K+1){

        ll ans = (nCk(r+1, i, MOD) * nCk(b-1, i-1, MOD)) % MOD;

        print(ans);

    }



    return 0;

}


// Optimized solution to the same problem (faster):
// abc132_d

#include <bits/stdc++.h>

#ifdef LOCAL

#include "../cxx-prettyprint/prettyprint.hpp"

#endif

using namespace std;



typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> P;



#define REP(i, n) for (int (i) = 0 ; (i) < (int)(n) ; ++(i))

#define REPN(i, m, n) for (int (i) = m ; (i) < (int)(n) ; ++(i))

#define REP_REV(i, n) for (int (i) = (int)(n) - 1 ; (i) >= 0 ; --(i))

#define REPN_REV(i, m, n) for (int (i) = (int)(n) - 1 ; (i) >= m ; --(i))

#define ALL(x) x.begin(), x.end()



#define INF ((1 << 29)-1)

#define MOD (1000000007)



#define print2D(h, w, arr) REP(i, h) { REP(j, w) cout << arr[i][j] << " "; cout << endl; }

template<class T> void print(const T& x){cout << x << endl;}

template<class T, class... A> void print(const T& first, const A&... rest) { cout << first << " "; print(rest...); }

struct PreMain {PreMain(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(20);}} premain;





int mod_pow(ll a,ll b,int mo){int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}



ll factorial[2001];

void make_factorial(int mo){ factorial[0] = 1; REPN(i, 1, 2001) factorial[i] = (i * factorial[i-1]) % mo; }

ll get_factorial(int n){ return n >= 0 ? factorial[n] : 0; }



ll nCk(int n, int k, int mo){

    ll a = get_factorial(n);

    ll b = get_factorial(n-k) * get_factorial(k) % mo;

    ll b_inv = mod_pow(b, mo-2, mo);

    return a * b_inv % mo;

}



int main() {

#ifdef LOCAL

    ifstream in("../arg.txt"); cin.rdbuf(in.rdbuf());

#endif



    int N, K;

    cin >> N >> K;



    int b = K;

    int r = N - K;



    make_factorial(MOD);



    REPN(i, 1, K+1){

        ll ans = (nCk(r+1, i, MOD) * nCk(b-1, i-1, MOD)) % MOD;

        print(ans);

    }



    return 0;

}


// Fastest known solution for this problem: