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>
using namespace std;
#define ll long long                                    // long long省略
#define pb push_back                                    // push_back省略
#define mp make_pair                                    // make_pair省略
#define fi first                                        // first省略
#define se second                                       // second省略
#define itn int                                         // int誤字保険                                   // cout誤字保険
#define vecotr vector                                   // vector誤字保険
#define ednl endl                                       // endl誤字保険
#define opt() cin.tie(0); ios::sync_with_stdio(false)   // 入出力速度改善

#define rep(i,l,r) for(ll i=(l);i<(r);i++)               // 範囲[l, r)で刻み1のfor文(順方向)
#define repp(i,l,r,k) for(ll i=(l);i<(r);i+=(k))         // 範囲[l, r)で刻みkのfor文(順方向)
#define rrep(i,l,r) for(ll i=(r-1);i>=(l);i--)           // 範囲[l, r)で刻み1のfor文(逆方向)
#define rrepp(i,l,r,k) for(ll i=(r-1);i>=(l);i-=(k))     // 範囲[l, r)で刻みkのfor文(逆方向)
#define all(x) (x).begin(), (x).end()                   // vectorのポインタ位置指定用
#define max(p,q)((p)>(q)?(p):(q))                       // max拡張
#define min(p,q)((p)<(q)?(p):(q))                       // min拡張
#define bit(n,m)(((n)>>(m))&1)                          // 変数nのm番目のbitを取り出す

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 (b<a) { a=b; return 1; } return 0; }

int dy[]={0, 1, 0, -1};                                 // 4方向近傍
int dx[]={1, 0, -1, 0};                                 // 4方向近傍

#define INF ((1LL<<62)-(1LL<<31))
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p) pom(a,p-2,p)

////////////////////////////////////////////////////////////////////////////////////////
int main(){
	ll n, k; cin >> n >> k;
	double res = 0;
	// サイコロを振ってk以上が出た場合
	if(n >= k)
		res += double( (n - k + 1) ) / double( n );
  	//cout << res << endl;
	rep(i,1,min(n+1,k)){
		ll tmp = i, cnt = 0;
		while(tmp < k){
			tmp *= 2;
			cnt++;
		}
      	//cout << cnt << " " << pow(0.5, cnt) / n << ednl;
		res += pow(0.5, cnt) / n;
	}
	cout << fixed << setprecision(12) << res << endl;
}