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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;
use std::str::FromStr;

fn input_tuple() -> (u64, u64) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<u64> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1])
}

fn dru(n: u64, m: u64) -> u64 {
    if n % m == 0 {
        return n / m;
    } 
    n / m + 1
}

fn main() {
    let (n, k) = input_tuple();
    let mut boundaries: Vec<u64> = Vec::new();
    let mut i = 0u32;
    loop {        
        let boundary = dru(k, 2u64.pow(i));
        boundaries.push(boundary);
        if boundary <= 1 {
            break;
        }
        i += 1;
    }
    // println!("{:?}", boundaries);
    let mut pre = n + 1;
    let mut sum = 0f64;
    let mut dividor = 1f64;
    for boundary in boundaries {               
        if boundary <= n {
            // println!("({} - {}) / {} / {}", pre, boundary, dividor, n);
            sum += (pre - boundary) as f64 / dividor / n as f64;
            pre = boundary;            
        }
        dividor *= 2f64;       
    }
    println!("{}", sum);
}