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;
typedef long long ll;
const long long INF=1e18,MOD=1e9+7;
ll GCD(ll a,ll b){
    ll hoge;
    do{
        hoge=a%b;
        a=b;
        b=hoge;
    }while(hoge!=0);
    return a;
}
ll LCM(ll a,ll b){
    return a/GCD(a,b)*b;//overflow対策
}
int main(){
	int n,k;
	cin>>n>>k;
	vector<int>mother(n,n);
	for(int i=1;i<=n;i++){
		int hoge=i;
		while(hoge<k){
			hoge*=2;
			mother[i-1]*=2;
		}
	}
	ll lcm_num=mother[0];
	for(int i=0;i<n;i++)lcm_num=LCM(mother[i],lcm_num);
	ll son_sum=0;
	for(int i=0;i<n;i++)son_sum+=lcm_num/mother[i];
	double ans=(double)son_sum/lcm_num;
	printf("%.10f\n",ans);

}