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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i) 
#define PI acos(-1)
#define pcnt __builtin_popcountll
#define rng(a) a.begin(), a.end()
#define sz(x) (int)(x).size()
#define v(T) vector<T>
#define vv(T) v(v(T))
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;

template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) {if (a > b) { a = b; return true; } return false; }

ll INF = 1001001001;
ll LINF = 1001001001001001001ll;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, k;
    cin>>n>>k;

    double ans = 0;
    rep(i, n) {
        int d = i + 1;
        double cnt = 1;
        while (d < k) {
            cnt *= 0.5;
            d *= 2;
        }
        ans += cnt;
    }
    printf("%.15f\n", ans / n);
}