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>
#define ls(x) (x)<<1
#define rs(x) (x)<<1|1
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ll rt<<1
#define rr rt<<1|1
#define PI acos(-1)
#define MM int m=(l+r)>>1;
#define fu(i,r,t) for(int i=r;i<=t;i++)
#define fd(i,r,t) for(int i=r;i>=t;i--)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long
#define endl '\n'
#define lcm(a,b) (a*b)/__gcd(a,b)
#define Case(i,ans) cout<<"Case "<<i<<": "<<ans<<endl;
#define cp(i,ans) printf("%.if",ans);
#define cpp(i,ans) cout<<setprecision(i)<<fixed<<ans<<endl;
#define ppb pop_back
#define ppf pop_front
#define pb push_back
#define pf push_front
#define pq priority_queue
#define lowbit(x) (x)&(-x)
#define all(V) V.begin(),V.end()
#define ms multiset
#define mod(x) ((x+mo_num)%mo_num)
#define vc vector
#define vct vector<int>
#define fa(i,V) for(auto i:V)
#define out(i) cout<<(i)<<endl;
#define fi first
#define se second
#define fun(i) fu(i,1,n)
#define fut(i) fu(i,1,t)
#define fum(i) fu(i,1,m)
#define ld long double
//#define P pair<int,int>
using namespace std;
const int INF=LLONG_MAX;
const int maxn=1e6+5;
const int mo_num=1e9+7;
int n,m,a[maxn];
int Pow(int b) {
    int ans = 1, base = 2;
    while (b != 0) {
        if (b & 1 != 0)
            ans *= base;
        base *= base;
        b >>= 1;
    }
    return ans;
}
int32_t main()
{
    IOS
    cin>>n>>m;
    double ans;
    cout<<setprecision(12)<<fixed<<endl;
    fun(i)
    {
        int now=i,time=0;
        while(now<m)
        {
            now*=2;
            time++;
        }
        ans+=(double)1/(double)(Pow(time));
    }
    cout<<(double)ans/(double)n;
    return 0;
}