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.*;

public class Main {

  static Scanner sc = new Scanner(System.in);

  static long[] size;
  static long[] patty;

  public static void main(String[] args) {

    int N = sc.nextInt();
    long X = sc.nextLong();

    size = new long[N+1];
    patty = new long[N+1];

    size[0] = 1;
    patty[0] = 1;
    for (int i=1; i<=N; i++) {
      size[i] = size[i-1] * 2 + 3;
      patty[i] = patty[i-1] * 2 + 1;
    }

    System.out.println(f(N, X));
  }

  static long f(int n, long x) {
    if (n == 0) return 1;
    if (x == 1) return 0;
    if (x <= (size[n]-1)/2) {
      return f(n-1, x-1);
    } else if (x == (size[n]-1)/2 + 1) {
      return patty[n-1] + 1;
    } else {
      return patty[n-1] + 1 + f(n-1, x-(size[n]-1)/2-1);
    }
  }
  
}