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 static java.lang.Long.*;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
	public static long N;
	public static long X;

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		st = new StringTokenizer(br.readLine());
		N = parseLong(st.nextToken());
		X = parseLong(st.nextToken());

		long ans = pn(N,X);
		System.out.println(ans);

	}

	public static long pn(long n, long x) {
		if(n == 0 && x==1) {
			return 1;
		}else if (x == 1) {/*1*/
			return 0;
		} else if (1 < x && x <= 1 + total(n - 1)) {/*2*/
			return pn(n - 1, x - 1);
		} else if (x == 2 + total(n - 1)) {/*3*/
			return pn(n - 1, total(n - 1)) + 1;
		} else if (2 + total(n - 1) < x && x <= 2 + 2 * total(n - 1)) {/*4*/
			return pn(n - 1, total(n - 1)) + 1 + pn(n - 1, x - 2 - total(n - 1));
		} else {/*5*/
			return 2 * pn(n - 1, total(n - 1)) + 1;
		}

	}

	public static long total(long n) {
		if (n == 0)
			return 1;
		return 3 + total(n - 1) * 2;
	}
}