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 <algorithm>
#include <bitset>
#include <climits>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>

#define MOD 1000000007
#define MOD2 998244353
#define int long long
#define EPS 1e-9
//#define PI 3.14159265358979

#define rep(i, n) for (int i = 0; i < (int)(n); i++)

using namespace std;

template < typename T >
ostream &operator<<(ostream &os, const vector< T > &A) {
	for (int i = 0; i < A.size(); i++)
		os << A[i] << " ";
	os << endl;
	return os;
}
template <>
ostream &operator<<(ostream &os, const vector< vector< int > > &A) {
	int N = A.size();
	int M;
	if (N > 0)
		M = A[0].size();
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++)
			os << A[i][j] << " ";
		os << endl;
	}
	return os;
}

typedef pair< int, int > pii;
typedef long long ll;

struct edge {
	int from, to, d, c;
	edge(int _from = 0, int _to = 0, int _d = 0, int _c = 0) {
		from = _from;
		to = _to;
		d = _d;
		c = _c;
	}
	bool operator<(const edge &rhs) const {
		return (d == rhs.d) ? (c < rhs.c) : (d < rhs.d);
	}
};
struct aabb {
	int x1, y1, x2, y2;
	aabb(int x1, int y1, int x2, int y2) : x1(x1), y1(y1), x2(x2), y2(y2) {}
};

typedef vector< edge > edges;
typedef vector< edges > graph;
struct flow {
	int to, cap, rev, cost;
	flow(int to = 0, int cap = 0, int rev = 0, int cost = 0) : to(to), cap(cap), rev(rev), cost(cost) {}
};
typedef vector< vector< flow > > flows;

const int di[4] = {0, -1, 0, 1};
const int dj[4] = {-1, 0, 1, 0};
const int ci[5] = {0, 0, -1, 0, 1};
const int cj[5] = {0, -1, 0, 1, 0};
const ll LINF = LLONG_MAX / 2;
const int INF = INT_MAX / 2;
const double PI = acos(-1);

int pow2(int n) { return 1LL << n; }
template < typename T, typename U >
bool chmin(T &x, const U &y) {
	if (x > y) {
		x = y;
		return true;
	}
	return false;
}
template < typename T, typename U >
bool chmax(T &x, const U &y) {
	if (x < y) {
		x = y;
		return true;
	}
	return false;
}

struct initializer {
	initializer() {
		cout << fixed << setprecision(20);
	}
};
initializer _____;

int N, M, K, T, Q, H, W;
signed main() {
	cin >> N >> K;
	double ans = 0.0;
	rep(i, N) {
		int s = i + 1;
		double p = 1;
		while (s < K) {
			s *= 2;
			p *= 0.5;
		}
		ans += 1.0 / N * p;
	}
	cout << ans << endl;
	return 0;
}