Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

2 7

Sample Output 1

4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).

Sample Input 2

1 1

Sample Output 2

0

The bottom-most layer of a level-1 burger is a bun.

Sample Input 3

50 4321098765432109

Sample Output 3

2160549382716056

A level-50 burger is rather thick, to the extent that the number of its layers does not fit into a 32-bit integer.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

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();

		System.out.println(countP(n, x));

	}

	private static long countP(int n, long x) {
		long pCount = 0;

		if(n == 0) {
			// レベル0は肉一枚！
			return 1;
		}

		if(x <= n) {
			// 下からn枚は全部パン！
			return 0;
		}

		// 真ん中まで入ってるか
		long half = (long)Math.pow(2, n+1) - 2;
		if(x > half) {
			// 入ってる
			// 入ってるレベルの分をカウント
			pCount += (long)Math.pow(2,n) - 1;
			if(x != half) {
				// ピッタリじゃなければ
				// 真ん中のPをカウント
				pCount++;
				long nX = x - half -1;
				// 残りを再起呼び出し
				pCount += countP(n-1, nX);
			}
		} else {
			// 入ってない
			// レベルを下げてもう一回!
			// 一番下のパンの分を1枚引く
			return countP(n-1, x-1);
		}

		return pCount;

	}



}