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: Wrong Answer)

#include <stdio.h>

int main()
{
    long double a;
    int n,k,i,tmp,b=0,e=0;
    long double c,d=2; // c/d;
    scanf("%d %d",&n,&k);
    int x[100]={0};
    int y[100]={0};
    for(i=0;k>1;i++){
        x[i]=(k-1)/2+1;
        k=(k-1)/2+1;
        b++;
//        printf("[%d",x[i]);
    }
    for(i=0;i<b;i++){
        if(e>0){
            y[e] = x[i-1] - x[i];
//            printf("}%d",y[e]);
        }
        if(x[i] <= n){
            if(e==0){
                y[e]= n - x[i] + 1;
//                printf("{%d",y[e]);
            }
            e++;
        }
    }
    for(i=0;i<e;i++){
        for(int j=0;j<y[i];j++){
            tmp = 1;
            for(int k=e-i-1;k>0;k--){
                tmp*=2;
//                printf("]%d",tmp);
            }
            c += tmp;
//            printf(",%Lf",c);
        }
        d *= 2;
    }
//    printf("[%LF]",d);
    printf("%.12Lf",c/(d*3));
    
}