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: Accepted)

def bg_len(n)
  if n == 0
    1
  else
    3 + bg_len(n - 1) * 2
  end
end

def bg_p(n)
  if n == 0
    1
  else
    1 + bg_p(n - 1) * 2
  end
end

def bg_ans(n, x)
  if x == 0
    0
  elsif x == 1
    n == 0 ? 1 : 0
  elsif n == 0
    1
  else
    x -= 1
    l = bg_len(n - 1)

    if x < l
      bg_ans(n - 1, x)
    elsif x == l
      bg_p(n - 1)
    elsif x == (l + 1)
      bg_p(n - 1) + 1
    else
      p = bg_p(n - 1) + 1
      p + bg_ans(n - 1, x - l - 1)
    end
  end
end

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

puts bg_ans(n, x)