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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
 
class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        double N = Integer.parseInt(sc.next());
        double K = Double.parseDouble(sc.next());
        double sum = 0;
        if(N < K) {
            for(int i = 1; i <= N; i++) {
                int kijun = 0;
                if(Math.log(K / i) / Math.log(2) == (int)(Math.log(K / i) / Math.log(2))) {
                    kijun = (int)(Math.log(K / i) / Math.log(2));
                }else {
                    kijun = (int)(Math.log(K / i) / Math.log(2)) + 1;
                }
                sum += 1 / (double)N * Math.pow(0.5, kijun);
                System.out.println(sum);
            }
        }else {
            
            for(double i = 1; i < K; i++) {
                int kijun = 0;
                if(Math.log(K / i) / Math.log(2) == (int)(Math.log(K / i) / Math.log(2))) {
                    kijun = (int)(Math.log(K / i) / Math.log(2));
                }else {
                    kijun = (int)(Math.log(K / i) / Math.log(2)) + 1;
                }
                sum += (1 / N) * Math.pow(0.5, kijun);
            }
            
            sum += (1 / N) * (N - K + 1);
            System.out.println(sum);
        }
        
    }
}