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.Scanner;
import java.util.*;
public class Main {
    static long[] sum;
    static long[] p;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long x = sc.nextLong();
        sum = new long[n];
        p = new long[n];
        sum[0] = 1;
        p[0] = 1;
        for (int i = 1; i < n; i++) {
            sum[i] = 2 * sum[i - 1] + 3;
            p[i] = 2 * p[i - 1] + 1;
        }
        long ans = f(n, x);
        System.out.println(ans);
    }
    public static long f(int n, long x) {
        if (n == 0 && x == 0) {
            return 0;
        }
        if (n == 0 && x != 0) {
            return 1;
        }
        if (x <= 1) {
            return 0;
        } else if (x <= 1 + sum[n - 1]) {
            return f(n - 1, x - 1);
        } else if (x <= 2 + 2 * sum[n - 1]) {
            return p[n - 1] + 1 + f(n - 1, x - 2 - sum[n - 1]);
        } else {
            return 2 * p[n - 1] + 1;
        }
    }
}
