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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var
	N,K,i,p,q,x,tmp,get:Longint;
	ans,now,make:double;
begin
	read(N,K);
	if N>=K then begin
		ans:=(N-K+1)/N;
		q:=K-1;
	end
	else q:=N;
	make:=1;
	get:=1;
	for i:=1 to 4 do begin
		if q=0 then break;
		get:=get*2;
		make:=make*0.5;
		p:=K div get;
		if p<q then begin;
			ans:=ans+(q-p)/N*make;
			q:=p;
		end;
	end;
	for i:=q downto 1 do begin
		tmp:=(K+i-1)div i;
		x:=1;
		now:=1/N;
		while tmp>x do begin
			now:=now*0.5;
			x:=x*2;
		end;
		ans:=ans+now;
	end;
	writeln(ans:0:9);
end.