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 ll long long int
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define pp pop_back
#define in insert
#define ff first
#define ss second
#define  show for(int j=0;j<v.size();j++) cout<<v[j]<<" ";
#define ld long double
#define e endl
#define bc __builtin_popcount
#define MOD 1000000007
#define INT_SIZE 32
#define me(x) memset(x,0,sizeof(x))
ll fast_expo(ll x,ll p)
{if(p==0) return 1;
else if(p%2==0){ ll t=fast_expo(x,p/2)%MOD;return (t*t)%MOD;} else return (x*(fast_expo(x,p-1))%MOD)%MOD;}
int main()
{
    int n,k,j,i;
    cin>>n>>k;
    double ans,res=0;
    for(i=1;i<=n;i++)
    {
        ans=1.0/n;
        j=i;
      while(j<k)
      {
          j*=2;
          ans/=2;
      }
      res+=ans;
    }
    cout<<fixed<<setprecision(10)<<res<<e;
}