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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<vector>
#include<iostream>
#include<string>
#include<cstdlib>
using namespace std;
int main(void)
{
	long long int i, n, k,t;
	double sum=0;
	

	cin >> n >> k;
	vector<double> num(n,1.0);
	vector<long long int> mt(n, 0);
	for (i = 0; i < n; i++)
	{
		mt[i] = i + 1;
		t = 0;
		for (t = 0; mt[i] < k; t++)
		{
		
			mt[i] = mt[i] * 2;
		}
		mt[i] = t;
	}
	for (i = 0; i <n; i++)
	{
		for (t = 0; t < mt[i]; t++)
		{
			num[i] = num[i] / 2.0;
		}
		if (num[i] == 0)
		{
			num[i] = 1;
		}
		num[i] = num[i] / n;
	}

	for (i = 0; i < n; i++)
	{
		sum = sum + num[i];
	}
 
	printf("%.11lf\n", sum);



	return 0;
}