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: Accepted)

import java.util.*;
import java.util.stream.*;

public class Main {

	static long[] count = new long[51];
	static long[] pa = new long[51];

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		long X = in.nextLong();
		count[0] = pa[0] = 1L;
		for (int i = 1; i <= 50; i++) {
			count[i] = 3L + count[i - 1] * 2L;
			pa[i] = 1L + pa[i - 1] * 2L;
		}
		System.out.println(count(N, X));
	}

	public static long count(int n, long x) {
		if (n == 0) {
			return 1;
		}
		if (x == 1) {
			return 0;
		}

		long ans = 0;

		// 一番した食べる
		x--;

		// 下のN-1次元バーガー
		if (x <= count[n - 1]) {
			return count(n - 1, x);
		} else {
			ans += pa[n - 1];
			x -= count[n - 1];
		}

		if(x < 2) {
			return ans + x;
		}
		// まだ残りがある場合はこの次元の真ん中のP
		ans++;
		x--;
		// 上のN-1次元バーガー
		if (x <= count[n - 1]) {
			return ans + count(n - 1, x);
		}
		return ans + pa[n - 1];
	}
}