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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Set;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String str[] = br.readLine().split(" ");

		int n = Integer.parseInt(str[0]);
		long x = Long.parseLong(str[1]);

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

	static long solve(int n, long x) {
		if (n == 0)
			return 1;
		long len = (1L << (n + 1)) - 3;
		long num = (1L << n) - 1;
		if (x == 1)
			return 0;
		else if (x <= len + 1)
			return solve(n - 1, x - 1);
		else if (x == len + 2)
			return num + 1;
		else if (x <= (len + 1) * 2)
			return num + 1 + solve(n - 1, x - len - 2);
		else
			return num * 2 + 1;
	}
}