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 pb push_back
#define pii pair<int,int>
#define pll pair<ll, ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define ppi pair<pii,int>
#define pip pair<int,pii>
#define pdd pair<double, double>
#define f first
#define s second
#define MOD 1000000007
#define mkp make_pair
#define M_PI 3.14159265358979323846
#define FOR(i,l,r) for (int i=l;i<=r;i++)
#define LOR(i,l,r) for (ll i=l;i<=r;i++)
#define FORD(i,r,l) for (int i=r;i>=l;i--)
#define LORD(i,r,l) for (ll i=r;i>=l;i--)
#define INF 1000000000
#define CL(x) memset(x,0,sizeof(x))
#define DEB(x) cout << #x << " : " << x << '\n'
#define ALL(x) x.begin(), x.end()
#define SZ(x) x.size()
#define UI(x) (int)(x-'A')
#define LI(x) (int)(x-'a')
typedef long long ll;

int N,K;
double ans;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin>>N>>K;
    FOR( i,1,N )
    {
        double p = 1/(double)N;
        int n = i;
        while( n<K )
        {
            p/=2;
            n*=2;
        }
        ans+=p;
    }
    cout<<fixed<<setprecision(7)<<ans<<'\n';

    return 0;
}