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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

template<typename T>
T power(T a, T b) {
	if (b == 0) return 1;
	T res = power(a * a, b / 2);
	if (b % 2) res *= a;
	return res;
}

int main() {
	double N, K;
	cin >> N >> K;
	vector<int> a;
	for (double i = 1; i <= N; ++i) {
		double j = i;
		int b = 0;
		while (j < K) {
			j *= 2;
			++b;
		}
		a.push_back(b);
	}
	vector<pair<int, int>> c;
	int tmp = a[0], sum = 1;
	for (int i = 1; i < a.size(); ++i) {
		if (tmp != a[i]) {
			c.emplace_back(tmp, sum);
			tmp = a[i];
			sum = 1;
		}
		else ++sum;
	}
	if (sum) c.emplace_back(tmp, sum);
	for (int i = 0; i < a.size(); ++i) cout << a[i] << ' ';
	cout << endl;
	for (int i = 0; i < c.size(); ++i) cout << c[i].first << ' ' << c[i].second << ' ';
	double ans = 0.0;
	for (pair<int, int>& i : c) {
		ans += i.second * pow(0.5, i.first);
		// ans += pow(0.5, i);
	}
	ans *= (1.0 / N);
	printf("%.20f", ans);
}