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 <cstdlib>  // abs() for integer 絶対値求めやつ
#include <cmath>    // abs() for float, and fabs()
#include <algorithm>
#include <vector>
#include <string>
#include<iomanip>//浮動小数点数を出力する制度の指定
//使い方　数値をnumとして　cout<<setprecision(n)<<num ;nは桁数
#define rep(i,n) for(int i = 0; i < (n);i++)
#define SORT(a) sort((a).begin(),(a).end())
#define REV(a) reverse((a).begin(),(a).end())
using namespace std;
using ll =long long;
using db = double;
#define INF 2000000000
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
#define che(a,string) cout<<"//"<<string <<"=="<<(a)<<"//" <<endl;
bool IsInt(double a){int b = a/1;if(a==b){return true;}else {return false;}}
/*覚えてないことメaモ
  数値型から文字列に変換　to_string(number);これは簡単
  文字列から数値型に変換　stoi(number)
  文字列からllに変換　stoll(number)
  __gcd(a,b)は最大公約数を取ってくれる、aとb
  の大きい方をNとしたとき、計算量は√N
 */
const int mod = 1e9 + 7;
//ここから書き始める


//関数の定義はここ

int main(){
	cout << setprecision(20);
	db n, k;
	cin >> n >> k;
	db ans = 0;
	db coin = 0 ;
	for(int i = 1; i <= n; i++){
		db p = 1; 
		int x = i ;
		while( x < k){
			p *= 0.5;
			x *= 2;
		}
		ans += p / n;
	}
	cout << ans << endl;
}