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<stdio.h>
#include<iostream>
#include<string>
#include<math.h>
using namespace std;


void insertSort(double a[], int size) {

	int i, j;
	for (i = 1; i < size; i++) {
		j = i;
		while ((j > 0) && (a[j - 1] > a[j])) {
			swap(a[j - 1], a[j]);
			j--;
		}
	
	}
}



int main() {


	int n,k;
	cin >> n;
	cin >> k;
	
	int score = k;
	int count;

	
	double ans[100001];

	int i;
	for (i=1;i<=n;i++){
		score = i;
		count = 0;
		while (1) {
			if (score < k) {
				score *= 2;
				count++;
			}
			else
				break;
		}
		//p += pow(0.5, count) / (double)n;
		ans[i] = pow(0.5, count) / (double)n;
		//insertSort(ans, i);

		
	}
	double p = 0;
	for (i = 1; i <=n ; i++) {

		p += ans[i];
	}


	printf("%0.11lf\n", p);

	
	return 0;
}