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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>

int main(void){
  int i;
  int n, k;
  int m;//pointを何倍するとk以上になるか
  int mul;//ポイントを2^n倍した物
  double kakuritu=0;//確率

  scanf("%d %d", &n, &k);

  for(i=1; i<=n; i++){
    //ココでのiは最初の得点の値になる
    m=0;
    mul=i;
    while(k>mul){
      m++;
      mul*=2;
    }
    // printf("i=%d (1/n)*pow(1/2,m)=%lf\n", i,(1./n)*pow(1./2,m));
    // printf("1/n=%lf pow(1/2,m)=%lf\n", 1./n, pow(1./2, m));
    // printf("i=%d n=%d m=%d\n", i, n, m);
    kakuritu+=pow(1./2,m);//最後に1/nする
    // printf("kakuritu=%lf\n", kakuritu);
  }
  printf("%.12lf\n", (1./n)*kakuritu);

  return 0;
}