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: Runtime Error)

n, x = STDIN.gets.split(' ').map(&:to_i)

# layers(l) = 'B' + layers(l-1) + 'P' + layers(l-1) + 'B'

class Burger
  def initialize(n)
    @layers = [1]
    @number_of_patties = [1]

    n.times do |i|
      @layers << 2 * @layers[i] + 3
      @number_of_patties << 2 * @number_of_patties[i] + 1
    end
  end

  def solve(n, x)
    if n == 0
      if x == 0
        return 0
      else
        return 1
      end
    end

    if x == 1
      return 0
    elsif x > 1 && x <= @layers[n-1] + 1
      return solve(n-1, x - 1)
    elsif x == @layers[n-1] + 2
      return @number_of_patties[n-1] + 1
    elsif x > @layers[n-1] + 2 && x < 2*@layers[n-1] + 2
      return @number_of_patties[n-1] + 1 + solve(n - 1, x - 1 - 1 - @layers[n - 1])
    elsif x == 2*@layers[n-1] + 3
      return 2 * @number_of_patties[n-1] + 1
    end
  end
end

burger = Burger.new(n)
p burger.solve(n, x)