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: Accepted)

import static java.lang.System.*;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(in);

        int n = sc.nextInt();
		int k = sc.nextInt();
		
		int[] mul = new int[20];
		double[] prob = new double[20];
		mul[0] = 1; prob[0] = 1;
		for(int i=1; i<20; i++) {
			mul[i] = mul[i-1] * 2;
			prob[i] = prob[i-1] / 2;
		}
		
		double ans = 0.0;
		for(int i=1; i<=n; i++) 
			for(int j=0; j<20; j++) 
				if(i * mul[j] >= k) {
					ans += prob[j];
					//if(i < 10) out.println(i + " " + j + " " + mul[j] + " " + prob[j]);
					break;
				}
			
		out.println(ans/n);
    }
}