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

public class Main {

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

		final long n = sc.nextInt();
		final long k = sc.nextInt();

		sc.close();

		final BigDecimal ndiv = BigDecimal.ONE.divide(BigDecimal.valueOf(n), 10, BigDecimal.ROUND_HALF_UP);
		final BigDecimal kdiv = BigDecimal.ONE.divide(BigDecimal.valueOf(2), 10, BigDecimal.ROUND_HALF_UP);
		BigDecimal result = BigDecimal.ZERO;

		for (long i = 1; i <= n; i++) {
			int pointCount = 0;
			long point = i;
			while (point < k) {
				pointCount++;
				point = point * 2;
			}
			result = result.add(ndiv.multiply(BigDecimal.valueOf(Math.pow(kdiv.doubleValue(), pointCount))));
		}

		System.out.print(result);

	}

}