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: Runtime Error)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val N = sc.nextInt()
    val X = sc.nextLong()
    // レベルnのハンバーガーの層数はf(n)で表され,f(n) = 2f(n-1) + 3, f(0) = 1である.
    // つまり, レベルNのハンバーガーの場合f(N)枚であり, このうちX層食べた時のパティの枚数を知りたい.

    // もし, X < f(N) / 2 + 1 の時, ハンバーガーの真ん中の途中まで食べているので,
    // レベル N - 1 のハンバーガーを X - 1 層食べた時の パティの枚数 が答えになる.

    // X == f(N) / 2 + 1 の時, ハンバーガーの真ん中まで食べているので,
    // レベルN - 1のハンバーガーに含まれるパティの数だけ食べていることになる.

    // X > f(n) / 2 + 1 の時, ハンバーガーの真ん中よりも食べているので,
    // レベル N - 1 のハンバーガーに含まれるパティの数 +
    // レベル N - 1 のハンバーガーを X - (f(N) / 2 + 1) 層食べた時の パティの枚数 が答えになる.

    // こうして再帰的に求めていけるが, レベル 0 の ハンバーガーに含まれるパティの数は 当然 1 である.

    def solve(): Long = {
      val f = new Array[Long](N+1)
      f(0) = 1
      var i = 1
      while (i <= N) {
        f(i) = 2 * f(i-1) + 3
        i += 1
      }

      val dp = new Array[Long](N+1)
      dp(0) = 1

      // 5 枚 ,3 枚目 が半分, 4枚まで食べたい時,
      def dfs(level: Int, X: Long): Long = {
        val center = f(level) / 2 + 1
        if (X >= 1 && X < center) {
          dfs(level - 1, X - 1)
        } else if (X >= center) {
          val miniBarger =
            if (dp(level - 1) != 0)
              dp(level - 1)
            else {
              dp(level - 1) = dfs(level - 1, f(level - 1))
              dp(level - 1)
            }
          if (X == center)
            miniBarger + 1
          else if (X > center && X < f(level))
            miniBarger + 1 + dfs(level - 1, X - center)
          else if (X == f(level))
            miniBarger + 1 + miniBarger
          else 0
        } else 0
      }

      dfs(N, X)
    }

    println(solve())
  }

}