#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" void solve(long long m1) { if (m1 < sqrt(mod)) { cout << "★" << m1 << "/" << 1 << endl; return; } long long Ax = 1e18; long long Ay = 1e18; long long m2 = 1; for (int i = 1; i <= 50000; ++i) { long long x = m1 * i%mod; long long y = m2 * i%mod; if (abs(Ax) > (x)) { Ax = x; Ay = y; } } m1 = mod - m1; for (int i = 1; i <= 50000; ++i) { long long x = m1 * i%mod; long long y = m2 * i%mod; if (abs(Ax) > x) { Ax = -1 * x; Ay = y; } } cout << "★" << Ax << "/" << Ay << endl; return; } int main() { /*rep(i, 6) { long long m; cin >> m; solve(m); } return 0;*/ 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 / K; long long pb = 2 % K; 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 vectorAns(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; }