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.*;
public class Main {

	static long[] a, p;
	
	static long dfs(int N, long X) {
		if(N == 0)
			return X <= 0 ? 0 : 1;
		else if(X <= a[N-1] + 1)
			return dfs(N-1, X-1);
		else
			return p[N-1] + 1 + dfs(N-1, X-a[N-1]-2);
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		long X = sc.nextLong();
		a = new long[N];
		p = new long[N];
		a[0] = p[0] = 1;
		for(int i = 1; i < N; i++) {
			a[i] = 2 * a[i-1] + 3;
			p[i] = 2 * p[i-1] + 1;
		}
		System.out.println(dfs(N, X));	
	}

}