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 {
	private static long[] all;
	private static long[] p;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long x = sc.nextLong();
		all = new long[n + 1];
		p = new long[n + 1];

		all[0] = 1;
		p[0] = 1;
		for (int i = 1; i <= n; i++) {
			all[i] = 2 * all[i - 1] + 3;
			p[i] = 2 * p[i - 1] + 1;
		}

		System.out.println(solve(n, x));
	}

	private static long solve(int n, long x) {
		if (x <= 0 || n < 0) return 0;
		if (n == 0) return 1;
		if (x == 1) return 0;

		if (x <= 1 + all[n - 1]) return solve(n - 1, x - 1);
		if (x == 2 + all[n - 1]) return p[n - 1] + 1;
		if (x <= 2 * all[n - 1] + 2) return p[n - 1] + 1 + solve(n - 1, x - 2 - all[n - 1]);
		return 2 * p[n - 1] + 1;
	}
}
