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<bits/stdc++.h>
using namespace std;
long long n,k,res;
vector<long long> v;
long long lowgbnum(long long x,long long y)
{
	const long long x1=x;
	const long long y1=y;
	while(y) 
	{
	    long long z=x%y;
	    x=y;
	    y=z;
	}
	return x1*y1/x;
}
int main()
{
	cin>>n>>k;
	for(long long i=1;i<=n;i++)
	{
		res=0;
		res+=n;
		long long tmp=i,z=1;
		while(tmp<k)
		{
			tmp*=2;
			z*=2;
		}
		res*=z;
		v.push_back(res);
	}
	long long tmp=1;
	long long ans=0;
	for(long long i=0;i<v.size();i++)
		tmp=lowgbnum(tmp,v[i]);
	for(long long i=0;i<v.size();i++)
		ans+=tmp/v[i];
	double u=(double)ans/tmp;
	printf("%0.12lf",u);
	return 0;
}