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 sc=new Scanner(System.in);
		int n = sc.nextInt();
		long x = sc.nextLong();
		long b = x;		//下からN層
		long ans = 0;		//答え
		for (int i = n; i >= 0; i--) {
			long h = height(i);		//総枚数
			long mid = (h + 1) / 2;		//真ん中
			if (b == mid) {		//真ん中にぶち当たったら、とりあえず、N-1のサイズのバーガーの肉全部と一番真ん中の肉を食べられる
				ans += patties(i - 1);
				ans++;
				break;
			} else if (b == 1) {		//一番下ならパンしか食って無いので自明に0枚
				break;
			} else if (b == h) {		//上から下までまるまる食べているのならNのパティの数を加算する
				ans += patties(i);
				break;
			} else if (b < mid) {		//中央値よりも小さい時
				b--;
			} else {			//中央値よりも大きかったら、その中央値を下限としてまた計算し直す
				ans += patties(i - 1);
				ans++;
				b -= mid;
			}
		}
		System.out.println(ans);
	}
	public static long height(int n) {
		return (1L << (n + 2)) - 3;
	}
	public static long patties(int n) {
		if (n < 0) {
			return 0;
		}
		return (1L << (n + 1)) - 1;
	}
}