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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define mp make_pair
#define pb(x) push_back(x)
#define vi vector<int>
#define vs vector<string>
#define vll vector<long long>
#define vc vector<char>
#define vs vector<string>
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pil pair<int, long long>
#define pli pair<long long, int>
#define pcc pair<char, char>
#define pdd pair<double, double>
#define mll map<long long, long long>
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
#define flsh fflush(stdout)
#define gcd __gcd
#define clr(x) memset(x, 0, sizeof(x))
#define mod 1000000007LL
#define mod2 998244353LL
#define INF 1000000000007
 

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> o_tree;


// void solve()
// {
//     string s;
//     cin>>s;
//     ll x1=s[0]-'a';
//     ll x2=s[1]-'a';
//     ll x3=s[2]-'a';
//     ll x4=s[3]-'a';
//     ll f1=0,f2=0;

     
    
    
// }


void solve()
{
    ll n,k;
    cin>>n>>k;

    ld res=0;
    ll i;
    for(i=1;i<=n;i++)
    {
        ld tans =i;
        ld ttans = 1.0/n;
        // ll di = floor(log2(n/tans));
        ll no = i;
        ll no2=0;
        while(no<k)
        {
             no*=2;
             no2++;
        }
        res +=1.0*pow(0.5,no2)*ttans;
    }
    cout<<res<<"\n";
}


int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);


if (fopen("input.txt","r" ))
  {
     freopen("input.txt", "r", stdin);
     freopen("output.txt", "w", stdout);
  }
cout<<setprecision(20);


ll t=1;


// cin>>t;


while(t--)
{
solve();
} 


return 0;
}