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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <map>
 
using namespace std;

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

    double N, K;

    cin >> N >> K;

    double check = 1.0;
    double check_second;
    double check_count = 0.0;

    double result;
    double total;

    while(true)
    {
        if(check >= K)
        {
            result = 1.0 / N / pow(2.0, check_count);
            break;
        }
        check *= 2.0;
        check_count++;
    }

    total += result;

    for(double i = 2.0; i <= N; i++)
    {
        check_second = pow(2.0, check_count - 1.0);
        if(i * check_second >= K)
        {
            check = check_second;
            check_count--;
            if(1.0 / check > 1.0)
            {
                result = 1.0 / N;
            } else {
                result = 1.0 / N / check;
            }
        }

        total += result;
    }

    printf("%.12f", total);
}