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;

public class Main {

	static long level[];
	static long p[];

	public static long burger(int n, long x) {
		if(n == 0) {
			if (x < 1)
				return 0;
			else return 1;
		}
		
		
		
		if (1 + level[n - 1] >= x)
			return burger(n - 1, x - 1);
		else 
			return 1 + p[n - 1] + burger(n - 1, x - 2 - level[n - 1]);
	}

	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);

		int n;
		long x;
		n = sc.nextInt();
		x = sc.nextLong();

		sc.close();

		level = new long[n + 1];
		p = new long[n + 1];

		level[0] = 1;
		p[0] = 1;

		for (int i = 1; i <= n; ++i) {
			level[i] = 2 * level[i - 1] + 3;
			p[i] = 2 * p[i - 1] + 1;
		}

		long count = burger(n, x);

		System.out.println(count);

	}
}
