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);
	// 整数の入力
	long n = sc.nextLong();
	long x = sc.nextLong();

	// 出力
	System.out.println(p(n,x));
    }

    private static long p(long n, long x){
	if(n == 0){
	    return 1;
	}

	long len_n1 = 4;
	for (long i = 1; i <= n-1; i++){
	    len_n1 *= 2;
	}
	len_n1 -= 3;

	if(x == 1){
	    return 0;
	}else if(x >= 2 && x <= 1 + len_n1){
	    return p(n -1, x -1);
	}else if(x == 2 + len_n1){
	    return p(n -1, len_n1) + 1;
	}else if(x >= 3 + len_n1 && x <= 2 + 2 * len_n1){
	    return p(n -1, len_n1) + 1 + p(n -1, x - len_n1 - 2);
	}else{
	    return p(n -1, len_n1) * 2 + 1;
	}
    }
}
