Problem:
Score : 400 points

Problem Statement

In some other world, today is Christmas.

Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:

A level-0 burger is a patty.

A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.

For example, a level-1 burger and a level-2 burger look like BPPPB and BBPPPBPBPPPBB (rotated 90 degrees), where B and P stands for a bun and a patty.

The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?

Constraints

1 \leq N \leq 50

1 \leq X \leq ( the total number of layers in a level-N burger )

N and X are integers.

Input

Input is given from Standard Input in the following format:

N X

Output

Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.

Sample Input 1

2 7

Sample Output 1

4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).

Sample Input 2

1 1

Sample Output 2

0

The bottom-most layer of a level-1 burger is a bun.

Sample Input 3

50 4321098765432109

Sample Output 3

2160549382716056

A level-50 burger is rather thick, to the extent that the number of its layers does not fit into a 32-bit integer.

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

def calc(n, count, k)
  if n == 0
    return count 
  else
    calc(n-1, count * 2 + k, k)
  end
end

def solve(n, k, total_p, total, ans)
  if k == 0
    return ans
  elsif k == (total - 1) / 2
    return ans + (total_p + 1) / 2
  elsif k == total - 1
    return ans + total_p + 1
  end
  if (total - 1) / 2 > k
    solve(n - 1, k , (total_p - 1) / 2, (total - 3) / 2, ans)
  else
    solve(n - 1, k - 1 - total_p , (total_p - 1) / 2, (total - 3) / 2, ans + (total_p + 1) / 2)
  end
end
  
n, k = gets.split.map(&:to_i)
p = calc(n, 1, 1)
t = calc(n, 1, 3)
puts solve(n, k - 1, p, t, 0)