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 {
	
	//ABC 115 d 

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

		long ans = rec(n,x);
		System.out.println(ans);

	}
	
	static long rec(int n , long x) {
		
		if(n == 0) {
			return 1;
		}
	
		long ren = (1L << n+1) - 3;		//レベルn-1の層の枚数
		long num = (1L << n) - 1 ;	 	//レベルn-1にあるPの数
		
		if(x == 1) {
			return 0;
		}
		
		if(x <= ren +1) {
			return rec(n-1,x-1); 
		}
		else if (x == ren +2) {
			return num + 1;
		}
		else if (x < 2 * ren + 3) {
			return num + 1 + rec(n-1,x-ren-2);
		}
		else {
			return 2 * num +1;
		}
	}
	
	// B|(n-1Level)|P|(n-1Level)|B
	//  	①		②		③

}
