Problem:
Score : 300 points

Problem Statement

Snuke has a fair N-sided die that shows the integers from 1 to N with equal probability and a fair coin. He will play the following game with them:

Throw the die. The current score is the result of the die.

As long as the score is between 1 and K-1 (inclusive), keep flipping the coin. The score is doubled each time the coin lands heads up, and the score becomes 0 if the coin lands tails up.

The game ends when the score becomes 0 or becomes K or above. Snuke wins if the score is K or above, and loses if the score is 0.

You are given N and K. Find the probability that Snuke wins the game.

Constraints

1 ≤ N ≤ 10^5

1 ≤ K ≤ 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the probability that Snuke wins the game. The output is considered correct when the absolute or relative error is at most 10^{-9}.

Sample Input 1

3 10

Sample Output 1

0.145833333333

If the die shows 1, Snuke needs to get four consecutive heads from four coin flips to obtain a score of 10 or above. The probability of this happening is \frac{1}{3} \times (\frac{1}{2})^4 = \frac{1}{48}.

If the die shows 2, Snuke needs to get three consecutive heads from three coin flips to obtain a score of 10 or above. The probability of this happening is \frac{1}{3} \times (\frac{1}{2})^3 = \frac{1}{24}.

If the die shows 3, Snuke needs to get two consecutive heads from two coin flips to obtain a score of 10 or above. The probability of this happening is \frac{1}{3} \times (\frac{1}{2})^2 = \frac{1}{12}.

Thus, the probability that Snuke wins is \frac{1}{48} + \frac{1}{24} + \frac{1}{12} = \frac{7}{48} \simeq 0.1458333333.

Sample Input 2

100000 5

Sample Output 2

0.999973749998

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define irep(i,n) for(int i = 0; i < (int)(n); i++)
#define irep2(i,a,n) for(int i = (int)(a); i <= (int)(n); i++)
#define irrep(i,n) for(int i = (int)(n-1); i > -1; i--)
#define irrep2(i,n,a) for(int i = (int)(n); i >= (int)(a); i--)

#define fi first
#define se second

using namespace std;

using ll = long long;
using v_int = vector<int>;
using v2_int = vector<v_int>;
using v_ll = vector<ll>;
using v2_ll = vector<v_ll>;
using v_string = vector<string>;
using v_bool = vector<bool>;
using v2_bool = vector<v_bool>;
using pii = pair<int, int>;
using mii = map<int, int>;

const double PI = 3.1415926535897932;
const int INF = (int)2e9;
const ll LINF = (ll)2e18;
const ll MOD = 1000000007;
const int dc[8] = {1, -1, 0,  0, 1, -1,  1, -1};
const int dr[8] = {0,  0, 1, -1, 1,  1, -1, -1};

template<typename T, typename U> inline
ostream& operator<<(ostream &o, pair<T,U> &p) {
    return o << "{" << p.first << "," << p.second << "}";
}
template<typename T> inline
istream& operator>>(istream &is, vector<T> &vec) {
    for(auto &v : vec) { is >> v; }
    return is;
}
template<typename T> inline
ostream& operator<<(ostream &os, vector<T> &vec) {
    for(auto &v : vec) { os << v << ",";}
    return os;
}
template<typename T> inline
ostream& operator<<(ostream &os, vector<vector<T> > &mat) {
    for(auto &row : mat) { os << row << endl; }
    return os;
}
template<typename T> inline
void chmin(T &a, T b) { a = min(a, b); }
template<typename T> inline
void chmax(T &a, T b) { a = max(a, b); }

int main(void)
{
    double n, k;
    cin >> n >> k;
    
    double ans;

    if(k <= n) {
        ans = (n-k+1)/n;
    } else {
        ans = 0.0;
    }
    
    for(int i = 1; i <= min(n, k-1); i++) {
        int j = i;
        double p = 1;
        while(j < k) {
            j *= 2;
            p /= 2.0;
        }
        ans += p/n;
    }
    printf("%.13f\n", ans);
    return 0;
} /*atcoder*/