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;
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pb push_back
#define pf push_front
#define F first
#define S second
#define ll long long
#define f(i) return cout<<i<<"\n",0;
#define vi vector<int>
#define vl vector<ll>
#define vii vector<pii>
#define vll vector<pll>
#define lp(i,a,b) for (int i=(a);i<=(int)(b);++i)
#define lp2(i,a,b) for (int i=(a);i>=(b);--i)
#define clr(c,x) memset(c,x,sizeof(c))
#define allr(v) v.rbegin(),v.rend()
#define all(v) v.begin(),v.end()
const int N=2e5+5,mod=(int)1e9+7;
int main()
{
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int k;
    long double n;
    cin>>n>>k;
    long double tot=0;
    lp(i,1,n)
    {
        long double prob=1.0/n;
        int score=i;
        while(score<k)
        {
            score*=2;
            prob*=0.5;
        }
        tot+=prob;
    }
    cout<<fixed<<setprecision(12)<<tot<<"\n";
    return 0;
}