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

}