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 REP(i, n) for(ll i = 0;i < n;i++)
#define REP1(i, n) for(ll i = 1;i < n;i++)
#define REPR(i, n) for(ll i = n-1;i >= 0;i--)
#define FOR(i, m, n) for(ll i = m;i < n;i++)

#define VSORT(v) sort(v.begin(), v.end())
#define VRSORT(v) sort(v.rbegin(), v.rend())

#define ll long long
#define ALL(a)  (a).begin(),(a).end()

#define pb(a) push_back(a)

int n,k;


int main(){
  double ans=0.0;
  cin >> n >> k;
  k--;
  REP(i,n){
    int ii=i+1;
    if (ii<=k){//コイン
      int flg=0;
      REP(j,31){
        if ((ii<<(j+1))>k){
          flg=1<<(j+1);
          break;
        }
      }
      ans+=1.0/flg;
      
    }else{
      ans+=1.0;
    }
  }
  ans/=n;



  cout<<setprecision(25)<<ans<<endl;
  return 0;
}