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 void main(String[] arg) {
        long N, X;
        Scanner scan = new Scanner(System.in);
        N = scan.nextLong();
        X = scan.nextLong();
        long[] a = new long[(int)N + 1];
        long[] p = new long[(int)N + 1];
        a[0] = 1;
        p[0] = 1;

        for (int i = 1; i <= N; i++) {
            a[i] = 2 * a[i - 1] + 3;
            p[i] = 2 * p[i - 1] + 1;
        }
        scan.close();
        System.out.println(f(N, X, a, p));
    }

    private static long f(long N, long X, long[] a, long[] p) {
        if (X == 1) {
            return N == 0 ? 1 : 0;
        }
        else if (X <= 1 + a[(int)N - 1]) {
            return f(N - 1, X - 1, a, p);
        }
        else if (X == 2 + a[(int)N - 1]) {
            return p[(int)N - 1] + 1;
        }
        else if (X <= 2 + 2 * a[(int)N - 1]) {
            return p[(int)N - 1] + 1 + f(N - 1, X - 2 - a[(int)N - 1], a, p);
        }
        else {
            return 2 * p[(int)N - 1] + 1;
        }
    }
}