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.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 burger = 1;
		for(int i = 0; i < N; i++) {
			burger = 3 + 2 * burger;
		}
		long ans = calc(burger, X);
		System.out.println(ans);
	}

	static long calc(long burger, long x) {
		long ans = 0;
		if(burger == 1) {
			ans = 1;
		} else {
			long tmp = (burger - 3) / 2;
			if(x < 2) {
				ans = 0;
			} else if(2 <= x && x < tmp + 2) {
				ans = calc(tmp, x - 1);
			} else if(tmp + 2 == x) {
				ans = calc(tmp, tmp);
				ans++;
			} else if(tmp + 2 < x && x < tmp * 2 + 3) {
				ans = calc(tmp, x - tmp - 2);
				ans += calc(tmp, tmp);
				ans++;
			} else if(tmp * 2 + 3 <= x) {
				ans = 2 * calc(tmp, tmp);
				ans++;
			}
		}
		return ans;
	}
}