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 {
	public static void main(String args[]) {
		Scanner scn = new Scanner(System.in);
		int N = scn.nextInt();
		long X = scn.nextLong();
		long[] A = new long[N + 1];
		long[] P = new long[N + 1];
		A[0] = 1;
		P[0] = 1;
		for(int i = 1;i <= N;i++) {
			A[i] = 2 * A[i-1] + 3;
			P[i] = 2 * P[i-1] + 1;
		}

		long ans = 0;
		int k = N;
		int n = N;
		while(X >0) {
			for(;k >= 0;k--) {
				if(X >= A[k] + (n-k)) {
					ans += P[k];
					X -= A[k] + (n - k);
					n = k;
					if(X > 0) {
						ans++;
						X--;
					}
					break;
				}
			}
			if(k == -1) {
				break;
			}
		}

		System.out.println(ans);

	}
}
