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: Runtime Error)

import java.util.Scanner;

public class Main {
    long patty[] = new long[51];
    long size[] = new long[51];

    public void main(Scanner sc) {
        int n = sc.nextInt();
        long x = sc.nextLong();

        patty[0] = 1;
        size[0] = 0;
        for (int i = 1; i <= 50; i++) {
            patty[i] = patty[i - 1] * 2 + 1;
            size[i] = patty[i] * 2 - 1;
        }

        System.out.println(calc(n, x));
    }

    private long calc(int level, long x) {
        if (x == 0) {
            return 0;
        }

        if (x * 2 > size[level]) {
            // 中央のパティより右
            return patty[level - 1] + 1 + calc(level - 1, x - size[level - 1] - 2);
        } else {
            // 中央のパティより左
            return calc(level - 1, x - 1);
        }
    }

    public static void main(String[] args) {
        try (Scanner sc = new Scanner(System.in)) {
            new Main().main(sc);
        } catch (Exception e) {
            throw e;
        }
    }
}