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