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>
#define mov(i,a,b) for(int i = a; i < b; i++)
#define MOD 1000000007
#define INF 1e9+5
typedef long long ll;
using namespace std;
bool isPowerOfTwo(ll n){
	return (ceil(log2(n)) == floor(log2(n)));
}
long long binpow(long long a, long long b){
	long long res = 1;
	while(b>0){
		if(b&1) res *= a;
		a = a*a;
		b >>= 1;
	}
	return res;
}
bool sortbysec(const pair<int,int> &a, 
              const pair<int,int> &b) 
{ 
	bool x = false;
	if((a.second*a.first) < b.second*b.first){
		x = true;
	}
    return x;
}
int main(){
	
	ll n,k;
	cin>>n>>k;
	double ans = 0.0;
	for(int i = 1; i <= n; i++){
		double inter = 0.0;
		inter = (1.0/(double)(n));
		int t = i;
		while(t < k){
			inter *= (double)(0.5);
			t = t*2;
;		}
		ans += inter;
	}
	printf("%.12f",ans);
}