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.

Examples

Input

2 7


Output

4


Input

1 1


Output

0


Input

50 4321098765432109


Output

2160549382716056
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    long X = scanner.nextLong();
    System.out.println(count(N, X));
  }

  private static long count(int n, long x) {
    if ((1L << (n + 2)) - 3 < x) throw new IllegalArgumentException();
    if (n == 0) return 1;
    if (x == 1) return 0;
    x--;
    if (x <= (1L << (n + 1)) - 3) return count(n - 1, x);
    x -= (1L << (n + 1)) - 3;
    if (x == 1) return 1L << n;
    x--;
    if (x <= (1L << (n + 1)) - 3) return count(n - 1, x) + (1L << n);
    return (1L << (n + 1)) - 1;
  }
}