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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define sws ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ll long long
#define vi vector<int>
#define pb push_back
#define pii pair<int, int>
#define mp make_pair
#define ff first
#define ss second
#define vvi vector<vi>
#define all(x) x.begin(), x.end()
#define fo(i, n) for( int i = 0; i < n; i++ )

const int INF = 1e9;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;


int main() {

    int n, k;
    cin >> n >> k;

    long double ans = 0;

    for ( int i = 1; i <= n; i++ ) {
        long double aux = 1.0 / n;
        int at = i;

        while ( at < k ) {
            at *= 2;
            aux *= 0.5;
        }
        ans += aux;
    }


    cout << fixed << setprecision(20) << ans << endl;

    return 0;
}