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 repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
 
//C++14 (GCC 5.4.1)
int main() {
    double N,K;
    cin >>N>>K;
    //-----------------------------------
    double ans=0;
    vector<int> Kd(30);
    Kd[0]=0;
    Kd[1]=1;
    int maxi;
 
    repd(i,1,29){
        Kd[i+1]=Kd[i]*2;
    }
    rep(i,N){
        rep(j,29){
            if((i+1)*Kd[j]<K && K<=(i+1)*Kd[j+1]){
                ans+=1.0/N/Kd[j+1];
                break;
            }
        }
    }
 

    printf("%.15lf\n",ans);

    return (0);
}