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[] a;
    static long[] p;
    
    public static void main(String[] args) throws Exception {
    
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        long X = sc.nextLong();
        
        a = new long[N+1];
        a[0] = 1L;
        p = new long[N+1];
        p[0] = 1L;
        
        for(int i = 1; i <= N; i++){
            a[i] = a[i-1] * 2 + 3;
            p[i] = p[i-1] * 2 + 1;
        }
        
        System.out.println( f(N, X) );
    }
    
    static long f(int N, long X){
        
        if( N == 0 ){
            return 1;
        }
        
        if( X == 1 ){
            return 0;
        }
        
        if( 1 < X && X <= a[N-1] + 1 ){
            return f(N-1, X-1);    
        }
        
        if( X == a[N-1] + 2 ){
            return p[N-1] + 1;
        }
        
        if( a[N-1] + 2 < X && X <= 2 * a[N-1] + 2 ){
            return p[N-1] + 1 + f(N-1, X-a[N-1]-2);
        }
        
        // X == a[N]
        return p[N];
    }
}
