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.Collections.Generic;
using System.Linq;
using System.Text;

namespace aojpro
{
    class Program
    {

        static long[] a;
        static long[] p;
      
        static long Paty(long N, long X)
        {
            if(N == 0)
            {
                if(X <= 0)
                {
                    return 0;
                }
                else
                {
                    return 1;
                }
            }
            else if(X > a[N - 1] + 1)
            {
                return Paty(N - 1, X - a[N - 1] - 2) + 1 + p[N - 1];
            }
            else
            {
                return Paty(N - 1, X - 1);
            }

        }


        static void Main(string[] args)
        {
            string[] str = Console.ReadLine().Split(' ');

            long n = long.Parse(str[0]);
            long x = long.Parse(str[1]);

            a = new long[n + 1];
            p = new long[n + 1];

            a[0] = 1;
            p[0] = 1;

            for(int i = 1; i <= n; i++)
            {
                a[i] = 2 * a[i - 1] + 3;
                p[i] = 2 * p[i - 1] + 1;
            }

            Console.WriteLine(Paty(n, x));

        }
    }
}