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.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	long p[];
	long b[];
	long a[];
	public void run() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long x = sc.nextLong();

		p = new long[51];
		b = new long[51];
		a = new long[51];
		p[0]=1;
		b[0]=0;
		a[0]=1;
		for(int i=0; i<50; i++) {
			p[i+1] = 2 * p[i] + 1;
			b[i+1] = 2 * b[i] + 2;
			a[i+1] = p[i+1]+b[i+1];
		}

		long ans = f(n, x);
		System.out.println(ans);
		sc.close();
	}

	long f(int n, long x) {
		if(n==0) {
			if(x<=0) return 0;
			else return 1;
		}
		if(a[n-1] +1 >= x) {
			return f(n-1, x-1);
		}
		else if(a[n-1] +2 == x) {
			return p[n-1] + 1;
		}
		else if( a[n-1] + 2 < x && x < a[n]) {
			return p[n-1] + 1 + f(n-1, x-a[n-1]-2);
		}
		else {
			return p[n];
		}
	}
}
