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)

#
# File    : D.rb
# Author  : Kazune Takahashi
# Created : 12/8/2018, 9:19:27 PM
# Powered by Visual Studio Code
#

n, x = gets.chomp.split(" ").map{|i| i.to_i}

@l = Array.new(51)
@q = Array.new(51)

@l[0] = 1
@q[0] = 1

for i in 1...51
  @l[i] = 3 + 2 * @l[i - 1]
  @q[i] = 1 + 2 * @q[i - 1]
end

def calc(y, k)
  if k == 0
    return y
  end
  if y <= 1
    return 0
  elsif y <= @l[k] / 2
    return calc(y - 1, k - 1)
  elsif y == @l[k] / 2 + 1
    return @q[k - 1] + 1
  elsif y <= @l[k] - 1
    return @q[k - 1] + 1 + calc(y - @l[k]/2 - 1, k - 1)
  else
    return 2 * @q[k - 1] + 1
  end
end

puts calc(x, n)