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 {
	static long[] a;
	static long[] p;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = Integer.parseInt(sc.next());
		long X = Long.parseLong(sc.next());

		a = new long [N+1];
		p = new long [N+1];

		a[0] = 1;
		p[0] = 1;

		for(int i=1;i<N;i++) {
			a[i] = a[i-1]*2+3;
			p[i] = p[i-1]*2+1;
		}

		System.out.println(f(N,X));


	}

	static long f(int N, long X) {
		if(N==0) {
			if(X <= 0) {return 0;
			}else { return 1;}
		}else {
			if(X <= 1) {return 0;
			}else if(1 < X && X < 2 + a[N-1]) {return f(N-1, X-1);
			}else if(X == 2 + a[N-1]) { return p[N-1] + 1;
			}else if(2+a[N-1] < X && X < 2*a[N-1]+3) {return p[N-1]+1+f(N-1, X-a[N-1]-2);
			}else {return 2*p[N-1]+1;
			}

		}
	}

}