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 {
  public static long[] p;
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    long x = sc.nextLong();
    p = new long[n + 1];
    p[0] = 1;
    for(int i = 1; i <= n; i++) {
      p[i] = 2 * p[i - 1] + 3;
    }
    System.out.println(func(n, x));
  }

  public static long func(int s, long t) {
    if(s == 0) return 1;
    if(t == 1) {
      return 0;
    } else if(t <= (1 + p[s - 1])) {
      return func(s - 1, t - 1);
    } else if(t == (2 + p[s - 1])) {
      return (1 + func(s - 1, p[s - 1]));
    } else if(t < (3 + (2 * p[s - 1]))) {
      return (1 + func(s - 1, p[s - 1]) + func(s - 1, t - (2 + p[s - 1])));   
    } else {
      return (2 * func(s - 1, p[s - 1]) + 1);  
    }
  }
}
