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 int n ;
    static long x;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        n = Integer.parseInt(sc.next());
        x = Long.parseLong(sc.next());
        System.out.println(dfs(x,n));
        sc.close();
    }

    public static long dfs(long remain,int level){
        if(level == 0){
            return 1;
        }else{
            if(remain==1){
                return 0;
            }else if(remain <=((1L<<(level+1))-2)){
                return dfs(remain-1,level-1);
            }else if(remain ==((1L<<(level+1))-1)){
                return 1L<<level;
            }else if(remain ==(1L<<(level+2))-3){
                return (1L<<(level+1))-1;
            }else {
                return (1L<<level) + dfs(remain-(1L<<(level+1))+1,level-1);
            }
        }
    }
}