#include <iostream>
using namespace std;

double d[2][50050];
int main() {
    int n;
    cin >> n;
    double p;
    cin >> p;
    d[1][1] = 1;
    for (int i = 2; i <= n; ++i) {
        double *from = d[~i & 1];
        double *to = d[i & 1];
        for (int j = 1; j <= i; ++j) {
            to[j] = 1 + from[j] * p + from[j - 1] * (1 - p);
        }
    }
    cout << fixed;
    cout.precision(10);
    for (int i = 1; i <= n; ++i) {
        cout << d[n & 1][i] << '\n';
    }
    
    return 0;
}