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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <map>
#include <cmath>
#include <iomanip>

using namespace std;
typedef long long ll;
int main() {
    ll N,K;
    cin >> N >> K;
    vector<long double> fl(N);
    vector<ll> ctm(N);
    long double ans=0;
    ll mact=0;
    for(int i=0;i<N;i++){
        ll ct=0;
        ll tmp=i+1;
        while(tmp<K){
            tmp*=2;
            ct++;
        }
        //fl[i]=(float)1.0/N*(float)1.0/pow(2,ct);
        ctm[i]=ct;
        mact=max(mact,ct);
        //ans+=fl[i];
    }
    ll tp=0;
    for(int i=0;i<N;i++){
        ll tmp=1;
        while(mact>ctm[i]){
            ctm[i]++;
            tmp*=2;
        }
        tp+=tmp;
    }
    ll bun=1;
    for(int i=0;i<mact;i++){
        bun*=2;
    }
    //ans=(float)1.0/N*tp/pow(2,mact);
    ans=(long double)tp/(long double)(N*bun);

    cout << fixed << setprecision(10) << ans << endl;
}