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>
#include <unistd.h>

#define fi first
#define se second
#define rep(i,n) for (ll i = 0; i < (n); ++i)
#define rep1(i,n) for (ll i = 1; i <= (n); ++i)
#define rrep(i,n) for (ll i = (n)-1; i >=0; --i)
#define ALL(a) a.begin(),a.end()
#define pb push_back
#define dame { puts("-1"); return 0;}
#define show(x) cerr<<#x<<" = "<<x<<endl;

using namespace std;
using ll = long long;
using ld = long double;
using pl = pair<ll, ll>;

typedef vector<ll> vl;
typedef vector<pl> vp;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }


// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const ll mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
} c(200005);

// Sieve of Eratosthenes
// https://youtu.be/UTVg7wzMWQc?t=2774
struct Sieve {
  int n;
  vector<int> f, primes;
  Sieve(int n=1):n(n), f(n+1) {
    f[0] = f[1] = -1;
    for (ll i = 2; i <= n; ++i) {
      if (f[i]) continue;
      primes.push_back(i);
      f[i] = i;
      for (ll j = i*i; j <= n; j += i) {
        if (!f[j]) f[j] = i;
      }
    }
  }
  bool isPrime(int x) { return f[x] == x;}
  vector<int> factorList(int x) {
    vector<int> res;
    while (x != 1) {
      res.push_back(f[x]);
      x /= f[x];
    }
    return res;
  }
  vector<pl> factor(int x) {
    vector<int> fl = factorList(x);
    if (fl.size() == 0) return {};
    vector<pl> res(1, pl(fl[0], 0));
    for (int p : fl) {
      if (res.back().first == p) {
        res.back().second++;
      } else {
        res.emplace_back(p, 1);
      }
    }
    return res;
  }
  vector<pair<ll,int>> factor(ll x) {
    vector<pair<ll,int>> res;
    for (int p : primes) {
      int y = 0;
      while (x%p == 0) x /= p, ++y;
      if (y != 0) res.emplace_back(p,y);
    }
    if (x != 1) res.emplace_back(x,1);
    return res;
  }
};

using data = pair<ll, vector<ll> >;
ll LINF = 1000000000000000000;

int main() {
  ll n,k;
  cin >> n >> k;

  ld ans = 0.0;

  vector<ll> v(n);
  rep(i, n) v[i] = i+1;
  
  ll tmp = n;
  ld last_p = 1.0;
  int last_pp = 0;
  if(n>=k) last_p = ld((k-1))/ld(n);

  vector<ld> z;
  vector<pair<ll, int> > y;
  while(true){
    rep(i, n){
      v[i] *= 2;
    }
    auto x = lower_bound(ALL(v), k) - v.begin();
    last_p *= 0.5;
    last_pp++;
    y.pb(make_pair(tmp-x, last_pp));
    z.pb( last_p * ld(tmp - x));
    tmp = x;
    if(x==0) break;
  }

  ll tmp_calc = 0;
  rep(i, z.size()) {
    tmp_calc += 1048576 / pow(2, y[i].se) * y[i].fi;
    ans += z[z.size()-i-1];
  }

  
  ans = ld(tmp_calc)/ ld(1048576) / ld(n);
  if(n>=k) ans *= ld((k-1))/ld(n);
  if(n>=k) ans += ld(n-k+1)/ld(n);

  
  cout << setprecision(15) << ans << endl;
  return 0;
}