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;
using ll = long long;
using itn = int;
using ld = long double;
template<class T>
using vec = vector<T>;

#define FOR(i,a,b) for(int i=((int)(a));i<((int)(b));i++)
#define REP(i,x) for(int i=0;i<(int)(x);i++)
#define all(x) (x).begin(),(x).end()


int main(){
  int n,k;
  cin >>n >>k;

  vec<ld> p(n);
  ld pa=0,pb=0;
  ld a;
  for(int i=1;i<=min(n,k);i++){
    ld b=1;
    a = log2((ld)(k/i)+1);
    REP(j,a) b /= 2;
    p[i-1] = b/(ld)(n);
  }
  REP(u,n) pa += p[u];

  if(n >= k) pb = (ld)(n-k+1)/(ld)(n);

  cout <<fixed <<setprecision(15) <<ld(pa+pb) <<endl;
}