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 PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php
#C - Dice and Coin

fscanf(STDIN, "%d %d", $N, $K);

$P = [];

#サイコロの目が$iの時、得点が$K以上になるためには、$coin_count連続で表が出る必要がある。
for ( $i = 1; $i <= $N; $i++ ) {

    #scoreが$K以上になるためのコイントス回数
    $score = $i;
    $coin_count = 0;
    while ( $K >= $score) {
        $score *= 2;
        $coin_count++;
    }

    #コイントス回数からサイコロが$iの時の分母を計算
    $denominator = 1;
    for ( $j = 1; $j <= $coin_count; $j++ ) {
        $denominator *= 2;
    }

    $score = $i;
    $coin_count = 0;
    
    $P[] = $N * $denominator;
}

$numerator = 0;
foreach ($P as $el) {
    $numerator += ($P[0] / $el);
}

$probability = ($numerator / $P[0]);

echo $probability;