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 <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <map>
#include <iomanip>
#define _overload3(_1, _2, _3, name, ...) name
#define _rep(i, n) repi(i, 0, n)
#define repi(i, a, b) for (ll i = (ll)(a); i < (ll)(b); ++i)
#define rep(...) _overload3(__VA_ARGS__, repi, _rep, )(__VA_ARGS__)
#define ll long long
#define lld long double
#define ALL(x) x.begin(), x.end()
#ifdef DEBUG
#define line() cerr << "[" << __LINE__ << "] ";
#define dump(i) cerr << #i ": " << i << " ";
#define dumpl(i) cerr << #i ": " << i << endl;
#else
#define line(i)
#define dump(i)
#define dumpl(i)
#endif
using namespace std;

int main()
{
    int n, k;
    cin >> n >> k;
    double ans = 0.0;
    rep(i, 1, n + 1)
    {
        int x = i;
        double c = 1.0;

        while (x < k)
        {
            //cerr << x << "," << n << endl;
            x *= 2;
            c /= 2;
        }
        // cerr << c << endl;
        ans += c;
    }
    cout << std::setprecision(18) << ans / (double)n << endl;
    return 0;
}