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.*;

class Main{
	static long[]l=new long[51];
	static long[]b=new long[51];
	
	public static void main(String[] $){
		Scanner s=new Scanner(System.in);
		l[0]=1;
		b[0]=1;
		for(int i=1;i<l.length;++i) {
			l[i]=l[i-1]*2+3;
			b[i]=b[i-1]*2+1;
		}
		System.out.println(f(s.nextInt(),s.nextLong()));
	}

	static long f(int i,long a){
		if(a==l[i])
			return b[i];
		if(a>l[i-1]+2)
			return b[i-1]+1+f(i-1,a-l[i-1]-2);
		if(a==l[i-1]+2)
			return b[i-1]+1;
		if(a>1)
			return f(i-1,a-1);
		return 0;
	}
}
