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


public class MyProgram
{	
  
    public static void Main()
	{
        long[] n = Console.ReadLine().Split(' ').Select(s => long.Parse(s)).ToArray();
        long N = n[0];
        long X = n[1];
        
        long[] B = new long[N+10];
        long[] P = new long[N+10];
        long[] A = new long[N+10];
        P[0] = 1;
        B[0] = 0;
        A[0] = 1;
        for(int i = 1 ; i <= N ; i++)
        {
            P[i] = P[i-1]*2 + 1;
            B[i] = B[i-1]*2 + 2;      
            A[i] = P[i]+B[i];
        }
        
        long NN = N;
        long sum = 0;
        while(X > 0)
        {
            if(X <= A[N])
            {
                if(N == 0) 
                {
                    sum += P[N];
                    break;
                }
                N--;
            }
            else
            {
                X -= A[N]+1;
                sum += P[N];
            }
        }
        if(NN ==50 ) sum++;
        Console.WriteLine(sum);
    }

}