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;
import java.util.stream.IntStream;

public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			long k = scanner.nextLong();
			long[] sum = new long[n + 1], patty = new long[n + 1];
			sum[0] = patty[0] = 1L;
			IntStream.rangeClosed(1, n).forEach(i -> {
				sum[i] = (sum[i - 1] << 1) + 3;
				patty[i] = (patty[i - 1] << 1) + 1;
			});
			System.out.println(getResult(n, k, sum, patty));
		}
	}

	/**
	 * パティは何枚あるかを計算する
	 * 
	 * @param n
	 * @param k
	 * @param sum   階層数の配列
	 * @param patty バティ総数の配列
	 * @return パティは何枚あるか
	 */
	private static long getResult(int n, long k, long[] sum, long[] patty) {
		if (k <= n) {
			// パティまで食べられない
			return 0L;
		} else if (k == sum[n]) {
			// すべて食べた場合
			return patty[n];
		} else if (k <= sum[n - 1] + 1) {
			// 前半までしか食べない場合
			return getResult(n - 1, k - 1, sum, patty);
		} else if (k == sum[n - 1] + 2) {
			// 前半＋真ん中のパティを食べる場合
			return patty[n - 1] + 1;
		} else {
			// 後半まで食べる場合
			return patty[n] - getResult(n - 1, sum[n] - k - 1, sum, patty);
		}
	}
}
