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: Accepted)

#include <algorithm>
#include <iostream>
#include <limits>
#include <numeric>
#include <string>
#include <type_traits>

#define rep(i, x, n) for(int i = x; i < n; i++)
using namespace std;
const long long INF = 1LL << 60;
double ruijo(double x, int n) {
    double res = 1;
    rep(i, 0, n) { res *= x; }
    return res;
}
int intruijo(int x, int n) {
    int res = 1;
    rep(i, 0, n) { res *= x; }
    return res;
}
int main() {
    int N, K;
    cin >> N >> K;
    double res = 0;
    double n1 = 1 / (double)N;
    int r = 1;
    for(int i = N; i > 0; i--) {
        // cout << "i:" << i << endl;
        if(i >= K) {
            res += n1;
            // cout << n1 << endl;
            continue;
        }

        while(i *intruijo(2, r) < K) {

            r++;
        }
        res += n1 * ruijo(0.5, r);
        // cout << "r:"<<r<<"res:"<< n1 * ruijo(0.5, r) << endl;
    }
    // cout << res<< endl;
    printf("%.13f\n", res);

}