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

  def solve(): Unit = {
    val sc = new java.util.Scanner(System.in)
    val n = sc.nextInt()
    val x = sc.nextLong()

    val a = new Array[Long](n + 1)
    val p = new Array[Long](n + 1)
    a(0) = 1
    p(0) = 1
    for (i <- 1 until n + 1) {
      a(i) = 2 * a(i - 1) + 3
      p(i) = 2 * p(i - 1) + 1
    }

    def f(n: Int, x: Long): Long = {
      if (n == 0) {
        if (x == 0) 0L else 1L
      } else if (x == 1) {
        0L
      } else if (1 < x && x <= 1 + a(n - 1)) {
        f(n - 1, x - 1)
      } else if (x == 2 + a(n - 1)) {
        p(n - 1) + 1
      } else if (2 + a(n - 1) < x && x <= 2 + 2 * a(n - 1)) {
        p(n - 1) + 1 + f(n - 1, x - 2 - a(n - 1))
      } else {
        2 * p(n - 1) + 1
      }
    }

    println(f(n, x))

  }
}