#include<bits/stdc++.h>
using namespace std;
//type
typedef long long ll;
#define pii pair<int, int>
#define vi vector<int>
//x * y * 1.0 can cause overflow
//constant
#define inf (int)(1e9+7)
#define mod (ll)(1e9+7)
#define eps 1e-10
//omission
#define eb emplace_back
#define F first
#define S second
#define SZ(x) ((int)(x).size())
#define all(v) v.begin(), v.end()
#define ios cin.tie(0); ios::sync_with_stdio(false)
#define FIX(a) cout << fixed << setprecision(a)
//manip
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define UNIQUE(v) v.erase(unique(v.begin(), v.end(), v.end());
#define fill(x, y) memset(x, y, sizeof(x))
#define ceil(a, b) a / b + !!(a % b)
template<class T> T power(T a, T b) {return b ? power(a * a % inf, b / 2) * (b % 2 ? a : 1) % inf : 1;}
#define LB(v, x) (int)(lower_bound(v.begin(), v.end(), x) - v.begin())
#define UB(v, x) (int)(upper_bound(v.begin(), v.end(), x) - v.begin())
template<typename T> T gcd(T a, T b) {if (b == 0) return a; return gcd(b, a % b);}
template<typename T> T lcm(T a, T b) {return a / gcd(max(a, b), min(a, b)) * b;} //先に割る!!
#define outa(a, n) rep(i, n) cout << a[i] << " "; cout << endl
#define out(a) cout << a << endl
#define outv(v) rep(i, SZ(v)) cout << v[i] << " "; cout << endl
//loop
#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define rep3(i, st, n) for (int i = st; i < n; ++i)
//algorithm
//double pointer, l start, how many adds, can be 0 -> init r = l, sum = 0
//bfs, not x, y, i, j
//not !(i % 2), i % 2 == 0
//minimal required terms
//cast caution
//look constraints always

const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main() {
    //out(comb(3, 2));
    
    
    COMinit();
    
    int n, k; cin >> n >> k;
    if (n == 1) {
        //k == 1 b
        out(1);
        return 0;
    } else if (n == 2) {
        if (k == 1) {
            out(2);
        } else {
            out(1);
            out(0);
        }
        return 0;
    }
    
    
    ll ans = 0;
    out(n - k + 1);
    rep3(i, 2, k + 1) {
        //n+1箇所にi+1個の仕切り
        ans = COM(k - 1, i - 1) * (COM(n - k - 1, i) + COM(n - k - 1, i - 2) + 2 * COM(n - k - 1, i - 1));
        ans %= mod;
        out(ans);
        ans = 0;
    }
    
    
}

/*
 連続してるの圧縮して考えよう
 iに関して独立に考えていい
 
 i = 1
 n - k + 1
 
 
 i = 2
 b, r, b, r
 r, b, r, b
 
 i = 3
 b, r, b, r, b
 r, b, r, b, r
 
 
 i = 100
 b, r, b, r, b, r, ...
 
 10^9 + 7超える時点で再帰不能
 
 
 
 
 n個の玉並べる
 仕切りを入れる
 
 i = 1
 仕切り2つ
 
 違うは
 
 
 
 連続青を1まとめとする
 i = 2
 2箇所に分ける
 r, b, r, b, r
 端のr = 0も可
 
 i = 5
 r, b, r, b, r, b, r, b, r, b, r
 間のrは必ず1以上
 
 bに関してはすべて1個以上
 k個の間(k-1)i-1個選ぶ
 
 
 n - k個
 1. 両端ない
 2. 両端ある
 
 2. (n-k)個の間(n-k-1)i個選ぶ
 
 1. n-k個 間i-2箇所
 
 3. 片端
 n-k-1個からi-1個選ぶ
 */


Measured memory_bytes: 12160