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

i = 1
all = [1]
pati = [1]

n.times do
  all.push(all.last * 2 + 3)
  pati.push(1 + pati.last * 2)
end

all.reverse!
pati.reverse!
#p ['all', all]
#p ['pati', pati]

ans = 0
all.size.times do |i|
  # print [i, '', x, all[i], '', ans]
  if x <= 1
    ans += 1 if all[i] < 5
    break
  elsif x < 1 + all[i]
    x-=1
  elsif x == 1 + all[i]
    x = 0
    ans += pati[i]
  elsif x == 1 + all[i] + 1
    x = 0
    ans += pati[i] + 1
  elsif x < all[i]*2 + 2
    x -= (all[i] + 2)
    ans += pati[i] + 1
  elsif all[i]*2 + 2 == x
    x = 0
    ans += pati[i] * 2 + 1
  elsif all[i]*2 + 3 == x
    x = 0
    ans += pati[i] * 2 + 1
  else
    raise RuntimeError
  end
  # p ['->', x, ans]
end

# puts pati.inject(:+)
puts ans