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 rep(i, n) for(int i = 0, i##_len = (n); i < i##_len; ++i)
#define rep2(i, x, n) for(int i = x, i##_len = (n); i < i##_len; ++i)
#define all(n) begin(n), end(n)
using ll = long long;
using P = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<string>;
using vc = vector<char>;
using vb = vector<bool>;
vi dir = {-1, 0, 1, 0, -1, -1, 1, 1, -1};

int main() {
  double n, k;
  cin >> n >> k;
  double ans = 0;
  if(n > k) ans += (n - k + 1) / n;
  rep2(i, 1, min(n + 1, k)) {
    int num = 0;
    int j = i;
    while(j < k) {
      num++;
      j *= 2;
    }
    ans += 1 / pow(2, num) / n;
  }
  printf("%.12f\n", ans);
}