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)

import scala.io.StdIn.readLine

object Main {

  def main(args: Array[String]): Unit = {

    val Array(n, x) = readLine().split(' ').map(_.toLong)
    val ls: Vector[Long] = (1 to n.toInt)
      .foldLeft(Vector(1): Vector[Long])((acc, i) =>
        acc :+ (acc(i - 1) * 2 + 3))
    val ps: Vector[Long] = (1 to n.toInt)
      .foldLeft(Vector(1): Vector[Long])((acc, i) =>
        acc :+ (acc(i - 1) * 2 + 1))
    println(patty(n.toInt, x, ls, ps))

  }

  def patty(n: Int, x: Long, ls: Vector[Long], ps: Vector[Long]): Long = {

    if (x == 0) {
      0
    } else if (n == 0) {
      1
    } else if (x <= (ls(n) - 1) / 2) {
      patty(n - 1, x - 1, ls, ps)
    } else if (x == (ls(n) - 1) / 2) {
      patty(n - 1, x - 1, ls, ps) + 1
    } else {
      ps(n - 1) + 1 + patty(n - 1, x - ls(n - 1) - 2, ls, ps)
    }

  }

}