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