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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        D solver = new D();
        solver.solve(1, in, out);
        out.close();
    }

    static class D {
        long[] a;
        long[] p;

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int N = in.nextInt();
            long X = in.nextLong();
            a = new long[N + 1];
            p = new long[N + 1];
            a[0] = p[0] = 1;
            for (int i = 1; i <= N; i++) {
                a[i] = a[i - 1] * 2 + 3;
                p[i] = p[i - 1] * 2 + 1;
            }

            long count = calc(N, X);
            out.println(count);
        }

        private long calc(int N, long X) {
            if (X <= 1) {
                return N == 0 ? 1 : 0;
            } else if (X < 2 + a[N - 1]) {
                return calc(N - 1, X - 1);
            } else if (X == 2 + a[N - 1]) {
                return p[N - 1] + 1;
            } else if (X < 3 + 2 * a[N - 1]) {
                return p[N - 1] + 1 + calc(N - 1, X - 2 - a[N - 1]);
            }
            return 2 * p[N - 1] + 1;
        }

    }
}

