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)

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

$layer = {}
def burger_layer(level)
  return $layer[level] if $layer[level]
  return $layer[level] = 1 if level.zero?
  $layer[level] = 2 * burger_layer(level - 1) + 3
end

$p_count = {}
def p_count(level)
  return $p_count[level] if $p_count[level]
  return $p_count[level] = 1 if level.zero?
  $p_count[level] = 2 * p_count(level - 1) + 1
end

def ans(n, x)
  if x <= 1
    return n.zero? ? 1 : 0
  elsif x == burger_layer(n)
    return p_count(n)
  elsif x == burger_layer(n - 1)
    return p_count(n - 1)
  elsif x <= 1 + burger_layer(n - 1)
    return ans(n-1, x-1)
  elsif x == 2 + burger_layer(n - 1)
    return p_count(n - 1) + 1
  elsif x > 2 + burger_layer(n - 1)
    return ans(n-1, burger_layer(n-1)) + 1 + ans(n-1, x-burger_layer(n-1)-2)
  end
end

puts ans(n, x)