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 {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		long X = sc.nextLong();
		long burger = 1;
		for(int i = 0; i < N; i++) {
			burger = 3 + 2 * burger;
		}
		long ans = calc(burger, X);
		System.out.println(ans);
	}

	static long calc(long burger, long x) {
		long ans = 0;
		if(burger == 1) {
			ans = 1;
		} else {
			long tmp = (burger - 3) / 2;
			if(x < 2) {
				ans = 0;
			} else if(2 <= x && x < tmp + 2) {
				ans = calc(tmp, x - 1);
			} else if(tmp + 2 == x) {
				ans = calc(tmp, tmp);
				ans++;
			} else if(tmp + 2 < x && x < tmp * 2 + 3) {
				ans = calc(tmp, x - tmp - 2);
				ans += calc(tmp, tmp);
				ans++;
			} else if(tmp * 2 + 3 <= x) {
				ans = 2 * calc(tmp, tmp);
				ans++;
			}
		}
		return ans;
	}
}
