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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#ifdef ONLINE_JUDGE
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif
#define fu(a,b,c) for(int a=b;a<=(int)c;++a)
#define fd(a,b,c) for(int a=b;a>=(int)c;--a)
#define cu(a,b,c) for(int a=b;a<(int)c;++a)
#define cd(a,b,c) for(int a=b-1;a>=(int)c;--a)
#define gcd(a,b) __gcd(a,b) 
#define pii pair<int,int>
#define pll pair<long long,long long>
#define all(a) a.begin(),a.end()
#define F first
#define S second
#define pb push_back
#define pf push_front
#define MOD 1000000007 
#define MAXN 52
#define MAXH 20
#define MAXM 52
#define INF 100000000
#define ll long long
#define ull unsigned long long
#define RNG rng mt19937(time(NULL))
#define udist(a,b) uniform_int_distribution<uint32_t> distribution(a,b)
#define max3(a,b,c) max(a, max(b, c))
#define min3(a,b,c) min(a, min(b, c))
#define gmax(a,b) a = max(a, b)
#define gmin(a,b) a = min(a, b)
#define whatis(x) cerr << #x << " is " << x << endl;
#define fast_io() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define PI 3.1415926535897932384626433
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
int n,z;

int main(){
	fast_io();
	int n,k;
	cin >> n >> k;
	vector<int> weights(n + 1);
	fu(i,1,n){
		ll val = i;
		ll cnt = 0;
		while(val < k){
			++cnt;
			val<<=1;
		}
		weights[i] = cnt;
	}
	double ans = 0;
	fu(i,1,n){
		ans += (double)1/(1 << weights[i]);
	}
	ans/=n;
	cout.precision(13);
	cout << ans << '\n';
}