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 java.math.BigDecimal;
import java.util.Scanner;

public class Main {

	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int n = Integer.parseInt(sc.next());
		int k = Integer.parseInt(sc.next());

		double K = (double) k;

		//   double[] wariai = new double[n];

		double p = 0;
		double ans = 0;
		double tmp = 0;

		for (int i = 1; i <= n; i++) {
			p = (int) Math.ceil(Math.log(K / (double)i) / Math.log(2.0));
			//    	wariai[i]= Math.pow(0.5, p);
			if (p > 0) {
				tmp = Math.pow(0.5, p);
				ans += tmp;
			} else {
				//ans += (1 / n)*(n-i+1);
				ans += (n-k+1);
				break;
			}
		}
		BigDecimal Ans = BigDecimal.valueOf(ans/n);

		System.out.println(Ans);

	}

}