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 Scala (2.11.7) (judge verdict: Accepted)

object Main extends App {
  val in = new java.util.Scanner(System.in)
  val N = in.nextInt
  val X = in.nextLong

  val ls = (1 to N).scanLeft(1L)((z, i) => z * 2 + 3).toArray
  val ps = (1 to N).scanLeft(1L)((z, i) => z * 2 + 1).toArray

  def count(l: Int, eat: Long): Long = {
    if (l == 0) (if (eat <= 0) 0 else 1)
    else if (eat <= 1 + ls(l - 1)) count(l - 1, eat - 1)
    else ps(l - 1) + 1 + count(l - 1, eat - ls(l - 1) - 2)
  }

  println(count(N, X))
}