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{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        long X = sc.nextLong();
        long[] An = new long[N+1];
        long[] Bn = new long[N+1];
        long A = 5;
        long B = 3;
        for(int i = 1;i <= N;i++){
            An[i] = A;
            A = A*2+3;
        }
        for(int i = 1;i <= N;i++){
            Bn[i] = B;
            B = B*2+1;
        }
        System.out.println(f(N, X, An, Bn));
    }
    public static long f(int n, long x, long[] an, long[] bn){
        if(n == 1){
            if(x >= 5)  return 3;
            else    return Math.max(0, x-(long)1);
        }else if(an[n]-(long)1 <= x){
            return bn[n];
        }else if(an[n-1]+(long)2 <= x){
            return bn[n-1]+(long)1+f(n-1, x-(long)2-an[n-1], an, bn);
        }else{
            return f(n-1, x-(long)1, an, bn);
        }
    }
}