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 HashMap<String,Long> dic = new HashMap<>();
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long x = sc.nextLong();
        long layer = (long) (4*Math.pow(2,n)-3);
        dic.put(Long.toString(1),1L);
        long ans = helper(layer,x);
        System.out.println(ans);
    }
    // treat every sub-problem as a complete burger. Make-up for the lower buns when necessary
    static public long helper(long layer, long x){
        String k = Long.toString(layer)+"*"+Long.toString(x);
        long ans = 0;
        if(dic.containsKey(k)) return dic.get(k);
        if(x<=0||layer==0) return 0;
        if(layer==1) return 1;
        // in former step, the bun on the top may be annoying so that x>layer.
        // in case this happens, let x <= layer for all the time.
        if(x>layer) x=layer;
        if(2*x+1<=layer){
            ans += helper((layer-3)/2,x-1);
        } else{
            ans += helper((layer-3)/2,(layer-3)/2);
            ans += 1; // the patty in the middle
            if(x==layer) x--;// eating the top bun means nothing
            ans += helper((layer-3)/2,x-(layer+1)/2);
        }
        dic.put(k,ans);
        return ans;
    }
}