#include "bits/stdc++.h"
#include "atcoder/all"
using namespace std;
using namespace atcoder;
using mint = modint1000000007;
const int mod = 1000000007;
//using mint = modint998244353;
//const int mod = 998244353;
//const int INF = 1e9;
//const long long LINF = 1e18;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i,l,r)for(int i=(l);i<(r);++i)
#define rrep(i, n) for (int i = (n-1); i >= 0; --i)
#define rrep2(i,l,r)for(int i=(r-1);i>=(l);--i)
#define all(x) (x).begin(),(x).end()
#define allR(x) (x).rbegin(),(x).rend()
#define endl "\n"

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	//無限直線上で考える。
	int N, K;
	cin >> N >> K;
	int tmpK = K - 1;

	//2^(K-1) = aN + b
	mint pa = 2 / N;
	long long pb = 2 % N;
	mint a = 0;
	long long b = 1;
	while (tmpK > 0) {
		if (1 == tmpK % 2) {
			a = a * pa * N + pa * b + a * pb + b * pb / N;
			b = (b * pb) % N;
		}
		tmpK /= 2;
		pa = pa * pa * N + 2 * pa * pb + pb * pb / N;
		pb = (pb*pb) % N;
	}
	//cout << a.val() << "?" << b << endl;
	//周期がN
	vector<mint>Ans(N);
	rep(i, N) {
		int x = 2 * i + 1;
		int x1 = x % N;
		int x2 = (N - x1) % N;
		Ans[x1]++;
		Ans[x2]++;
	}
	rep(i, N) {Ans[i] *= a;}
	rep(i, b) {
		int x = 2 * i + 1;
		int x1 = x % N;
		int x2 = (N - x1) % N;
		Ans[x1]++;
		Ans[x2]++;
	}
	mint inv = 1;
	mint inv2 = (mint)1 / 2;
	while (K > 0) {
		if (1 == K % 2) {inv *= inv2;}
		K /= 2;
		inv2 *= inv2;
	}
	rep(i, N) {cout << (Ans[i] * inv).val() << endl;}
	return 0;
}