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 {

    private static int MOD = 1_000_000_007;

    public static void main(String[] args) {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }


    // int N = Integer.parseInt(scanner.nextLine());
    // int N = scanner.nextInt(), M = scanner.nextInt();
    // scanner.nextLine();
    // int[] n = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    // long[] n = Arrays.stream(scanner.nextLine().split(" ")).mapToLong(Long::parseLong).toArray();
    static long[] L;
    static long[] P;
    static void solve(Scanner scanner) {
        long[] n = Arrays.stream(scanner.nextLine().split(" ")).mapToLong(Long::parseLong).toArray();
        int N = (int) n[0];
        long X = n[1];
        L = new long[N + 1];
        P = new long[N + 1];
        L[0] = 1;
        P[0] = 1;
        for (int i = 1; i < N + 1; i++) {
            L[i] = 2 * L[i - 1] + 3;
            P[i] = 2 * P[i - 1] + 1;
        }

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

    static long dfs(int n, long x) {
        if (n == 0) {
            return (x > 0) ? 1 : 0;
        }
        if (x <= L[n - 1] + 1) {
            return dfs(n - 1, x - 1);
        }
        return P[n - 1] + 1 + dfs(n - 1, x - 2 - L[n - 1]);
    }

}
