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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;

namespace ABC115
{
    class Program
    {

        static long[] P;
        static long[] A;

        static void Main(string[] args)
        {
            long[] input = Console.ReadLine().Split().Select(long.Parse).ToArray();

            long N = input[0];
            long X = input[1];

            P = new long[N + 1];
            A = new long[N + 1];

            P[0] = 1;
            A[0] = 1;

            for (int i = 1; i < N; i++)
            {
                P[i] = 2 * P[i - 1] + 1;
                A[i] = 2 * A[i - 1] + 3;
            }
            Console.WriteLine(f(N, X));
        }

        static long f(long N, long X)
        {
            if (N == 0) { if (X <= 0) return 0; else return 1; }
            else if (X <= 1 + A[N - 1]) return f(N - 1, X - 1);
            else return P[N - 1] + 1 + f(N - 1, X - 2 - A[N - 1]);
        }
    }
}