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)

#define _CRT_SECURE_NO_WARNINGS

/* include ***********************/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

/* define *************************/
//for
#define REP(i,n) for(int i=0,i##_len=(int)(n);i<i##_len;i++)
#define REPS(i,n) for(int i=1,i##_len=(int)(n);i<=i##_len;i++)
#define RREP(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define RREPS(i,n) for(int i=(int)(n);i>0;i--)
#define FOR(i,s,n) for(int i=(int)(s),i##_len=(int)(n);i<i##_len;i++)
#define RFOR(i,s,n) for(int i=(int)(n)-1,i##_len=(int)(s);i>=i##_len;i--)

//printf
#define PRINTD(d) printf("%d\n",(d))
#define PRINTL(d) printf("%lld\n",(d))

//memset
#define m0(s) memset(s,0,sizeof(s))
#define ml(s) memset(s,63,sizeof(s))
#define fill(s,c) memset(s,c,sizeof(s))

#define INF 1e9
#define MOD 1000000007

typedef long long ll;
typedef unsigned long long ull;

int diff[4][2] = {
	{ 0, -1 },
	{ -1, 0 },
	{ 1, 0 },
	{ 0, 1 },
};
//今回の変数

int Min(int a, int b) {
	return (a)<(b) ? (a) : (b);
}
int Max(int  a, int  b) {
	return (a) > (b) ? (a) : (b);
}
ll Minl(ll a, ll b) {
	return (a)<(b) ? (a) : (b);
}
ll Maxl(ll a, ll b) {
	return (a)>(b) ? (a) : (b);
}

int main() {
	int n, k;
	scanf("%d %d", &n, &k);

	int dp[100001] = { 0 };
	RREPS(i, Max(n,k)) {
		if (i >= k) {
			dp[i] = 1;
			continue;
		}
		if (i * 2 >= k) {
			dp[i] = 2;
			continue;
		}
		dp[i] = dp[i * 2] * 2;
	}
	int max = 0;
	REPS(i, n) {
		max = Max(max, dp[i]);
	}
	
	ll num1 = 0;
	ll num2 = (ll)max * n;
	REPS(i, n) {
		num1 += max / dp[i];
	}
	double ans = num1 / (double)num2;
		printf("%.11lf\n", ans);
	
}