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Cow Camp

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Time limit
2000 ms
Memory limit
256 MB
Submissions
0
Correct
0
Solved by
0
AC rate

Statement

To qualify for cow camp, Bessie needs to earn a good score on the last problem of the USACOW Open contest. This problem has $T$ distinct test cases ($2\le T\le 10^3$) weighted equally, with the first test case being the sample case. Her final score will equal the number of test cases that her last submission passes.

Unfortunately, Bessie is way too tired to think about the problem, but since the answer to each test case is either "yes" or "no," she has a plan! Precisely, she decides to repeatedly submit the following nondeterministic solution:

if input == sample_input:
  print sample_output
else:
  print "yes" or "no" each with probability 1/2, independently for each test case

Note that for all test cases besides the sample, this program may produce a different output when resubmitted, so the number of test cases that it passes will vary.

Bessie knows that she cannot submit more than $K$ ($1\le K\le 10^9$) times in total because then she will certainly be disqualified. What is the maximum possible expected value of Bessie's final score, assuming that she follows the optimal strategy?

Input

The only line of input contains two space-separated integers $T$ and $K.$

Scoring

  • Test cases 3-6 satisfy $T\le 25$ and $K\le 100.$
  • Test cases 7-9 satisfy $K\le 10^6.$
  • Test cases 10-17 satisfy no additional constraints.

Output

The answer as a decimal that differs by at most $10^{-6}$ absolute or relative error from the actual answer.

Examples

Sample input 1
2 3
Sample output 1
1.875
Sample input 2
4 2
Sample output 2
2.8750000000000000000

Notes

In this example, Bessie should keep resubmitting until she has reached $3$ submissions or she receives full credit. Bessie will receive full credit with probability $\frac{7}{8}$ and half credit with probability $\frac{1}{8}$, so the expected value of Bessie's final score under this strategy is $\frac{7}{8}\cdot 2+\frac{1}{8}\cdot 1=\frac{15}{8}=1.875$. As we see from this formula, the expected value of Bessie's score can be calculated by taking the sum over $x$ of $p(x) \cdot x$, where $p(x)$ is the probability of receiving a score of $x$.

Here, Bessie should only submit twice if she passes fewer than $3$ test cases on her first try.

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