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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;

namespace Dice_and_Coin
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] X = Console.ReadLine().Split().Select(int.Parse).ToArray();
            int[] Y = new int[X[0]];
            double[] kakuritu = new double[X[0]];

            //何回コインを投げるか考える
            for(int i = 0; i < X[0]; i++)
            {
                int first = i + 1;          //さいころの目（１から）
                int count = 0;
                while (first <= X[1])
                {
                    first = first * 2;
                    count++;
                }
                Y[i] = count;
            }

            //各さいころの目で勝つ確率を計算
            for (int i = 0; i < X[0]; i++)
            {
                double tmp = 0.0;
                tmp = (1.0 / X[0]) * System.Math.Pow((1.0 / 2), Y[i]);
                kakuritu[i] = tmp;
            }

            //各さいころの目での勝つ確率を合計する
            double ans = 0.0;
            for (int i = 0; i < X[0]; i++)
            {
                ans = ans + kakuritu[i];
            }
            Console.WriteLine(ans);
        }
    }
}