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 = `dd`.split.map(&:to_i)
$ss = [1]
N.times{ $ss << $ss[-1]*2+3 }
$ps = [1]
N.times{ $ps << $ps[-1]*2+1 }

def dig(n,x)
  ans = case
  when n == 0; 1
  when x == 1; 0
  when x <= 1+$ss[n-1]; dig(n-1,x-1)
  when x == 1+$ss[n-1]+1; $ps[n-1]+1
  when x < $ss[n]; $ps[n-1]+1+dig(n-1,x-1-$ss[n-1]-1)
  when x == $ss[n]; $ps[n]
  end
  #p [n,x,$ss[n],$ps[n],ans]
  ans
end

p dig(N,X)