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;

class D
{
	static void Main()
	{
		var h = Array.ConvertAll(Console.ReadLine().Split(), long.Parse);
		long n = h[0], x = h[1];

		var p = new long[n + 1]; p[0] = 1;
		var c = new long[n + 1]; c[0] = 1;
		for (int i = 0; i < n; i++)
		{
			p[i + 1] = 2 * p[i] + 1;
			c[i + 1] = 2 * c[i] + 3;
		}
		Console.WriteLine(Find(n, x, p, c));
	}

	static long Find(long n, long x, long[] p, long[] c)
	{
		return x <= 1 ? 0 :
			x <= c[n - 1] + 1 ? Find(n - 1, x - 1, p, c) :
			x <= c[n - 1] + 2 ? p[n - 1] + 1 :
			x <= c[n] - 1 ? p[n - 1] + 1 + Find(n - 1, x - c[n - 1] - 2, p, c) :
			p[n];
	}
}