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.math.RoundingMode;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		System.out.println(new Main().doWork(args));
	}

	public String doWork(String[] args) {
		Scanner reader = new Scanner(System.in);

		int n = reader.nextInt();
		int k = reader.nextInt();

		reader.close();

		//		int rate = k / n;
		//		if(0 < (k % n)) {
		//			++rate;
		//		}

		BigDecimal ret = new BigDecimal(0);
		for (int i = 1; i <= n; ++i) {
//			int num = 1;
			int ct = 0;
			while (true) {
				if (k <= i * Math.pow(2, ct)) {
					break;
				}
				++ct;
			}

			BigDecimal wk = new BigDecimal(1);
			BigDecimal divisor = new BigDecimal(Math.pow(2, ct)).multiply(new BigDecimal(n));
			wk = wk.divide(divisor,200, RoundingMode.HALF_UP);
			ret = ret.add(wk);
//			return Double.toString(1.0d / Math.pow(2, ct));
		}

//		int num = 1;
//		int ct = 0;
//		while (true) {
//			if (k <= n * Math.pow(2, ct)) {
//				break;
//			}
//			++ct;
//		}

		return ret.toPlainString();
		//		return ch.substring(0, k - 1) + ch.substring(k - 1, k + 1 - 1).toLowerCase() + ch.substring(k + 1 - 1);
	}
}