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;

namespace abc115
{
    class MainClass
    {
        static long[] a;
        static long[] p;

        public static void Main(string[] args)
        {
            var nx = Console.ReadLine().Split();
            var n = long.Parse(nx[0]);
            var x = long.Parse(nx[1]);

            a = new long[n + 1];
            a[0] = 1;

            p = new long[n + 1];
            p[0] = 1;

            for (int i = 0; i < n; i++)
            {
                a[i + 1] = a[i] * 2 + 3;
                p[i + 1] = p[i] * 2 + 1;
            }

            var ans = m1(n, x);
            Console.WriteLine(ans);
        }

        static long m1(long n, long x)
        {
            if (x == 0) return 0;
            if (n == 0) return 1;

            var d = (a[n] + 1) / 2;
            if (x < d)
            {
                return m1(n - 1, x - 1);
            }
            else
            {
                return p[n - 1] + m1(n - 1, x - d) + 1;
            }
        }
    }
}