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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define FOR(i,n) for(int i=0;i<n;i++)
#define MOD 1000000007
typedef long long ll;
template <class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}else{return 0;}}
template <class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}else{return 0;}}
template <class T> inline T GCD(T a,T b){T c;while(b!=0){c=a%b;a=b;b=c;}return a;}
template <class T> inline T LCM(T a,T b){return a*b/GCD(a,b);}

int main(){
  long double n,k;
  cin >> n >> k;
  long double ans=0;
  long double num,cnt;
  for(int i=1;i<n+1;i++){
    if(i>=k){
      ans+=1/n;
    }else{
      num=i;
      cnt=0;
      while(k>num){
        num*=2;
        cnt++;
      }
      ans+=1/(pow(2,cnt)*n);
    }
  }

  cout << ans << endl;
  
}