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;



namespace CSharpSample01
{



    class Program
    {
        static long[] paty;
        static long[] layer;

        static long Solve(int n, long X)
        {
            
            if(X == 1)
            {
                if (n != 0) return 0;
                else return 1;
            }
            if (X < layer[n] / 2 + 1) return Solve(n - 1, X - 1);
            else if (X == layer[n] / 2 + 1) return paty[n - 1] + 1;
            else if (X < layer[n] - 1) return paty[n - 1] + 1 + Solve(n - 1, X - (layer[n] / 2 + 1));
            else return 2 * paty[n - 1] + 1;
        }

        static void Main(string[] args)
        {
            string[] input = Console.ReadLine().Split(' ');
            int N = int.Parse(input[0]);
            long X = long.Parse(input[1]);

            paty = new long[N + 1];
            paty[0] = 1;
            layer = new long[N + 1];
            layer[0] = 1;

            for(int i = 1; i <= N; i++)
            {
                layer[i] = 3 + 2 * layer[i - 1];
                paty[i] = 1 + 2 * paty[i - 1];
            }

            Console.WriteLine(Solve(N, X));
        }
    }
}