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: Wrong Answer)

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

namespace AtCoderCs
{
    class Program
    {
        static void Main()
        {

            var Arg = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();

            Console.WriteLine(MakingBurger(Arg[0], ref Arg[1]));

        }

        static long MakingBurger(long Level, ref long Num)
        {

            long Result = 0;

            if (Level == 0) return 1;

            // 1以下の場合
            if (Num <= 1)
            {
                Num = 0;
                return 0;
            }
            Result += MakingBurger(Level - 1, ref Num);

            switch (Num)
            {

                case 0:
                    return Result;
                    

                case 1:
                    Num = 0;
                    return Result + 1;

                default:

                    Result += MakingBurger(Level - 1, ref Num);
                    break;

            }

            if (Num == 0) return Result;
            else
            {

                Num--;
                return Result;

            }
        }


    }
}