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)

n,x = gets.chomp.split(" ").map(&:to_i)

class CalcBurger
  def initialize(i)
    @memo = {0 => 1 }
    @total = {
      0 => 1,
      1 => 3,
      2 => 7,
    }
    i.times do  |j|
      next if j == 0 
      @memo[j] = 2*@memo[j-1] + 3
      @total[j] = 2*@total[j-1] + 1
    end
  end

  def execute(i, x)
    if x == 1 
      0
    elsif 1 < x && x < @memo[i-1] + 1
      return execute(i-1, x-1)
    elsif x == 2 + @memo[i-1]
      return @total[i-1] + 1
    elsif 2 + @memo[i-1] < x && x <= 2 + 2 * @memo[i-1]
      return @total[i-1] + 1 + execute(i-1, x - 2 - @memo[i-1])
    elsif 3 + 2 * @memo[i-1]
      return @total[i-1]* 2 + 1
    end
  end
end
puts CalcBurger.new(n).execute(n, x)