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;
class Program
{
	static void Main(string[] args)
	{
		string[] str = Console.ReadLine().Split();
		int N = int.Parse(str[0]);
		long X = long.Parse(str[1]);
		long[]P = new long[51];
		long[]B = new long[51];
		P[0] = 1;
		B[0] = 0;
		for(var i=1;i<=50;i++){
			P[i] = P[i-1]*2+1;
			B[i] = B[i-1]*2+2;
		}
		long ans = fa(N,X,P,B);
		Console.WriteLine(ans);
	}
	public static long fa(int n,long x,long[]P,long[]B){
		long ans = 0;
		if(x<=1){
			ans = 0;
		} else if(x <= 1 + P[n-1] + B[n-1]){
			ans = fa(n-1,x-1,P,B);
		} else if(x <= 1 + P[n-1] + B[n-1] + 1){
			ans = P[n-1] + 1;
		} else if(x <= 1 + P[n-1] + B[n-1] + 1 + P[n-1] + B[n-1]){
			ans = P[n-1] + 1 + fa(n-1,x-1-P[n-1]-B[n-1]-1,P,B);
		} else {
			ans = P[n];
		}
		return ans;
	}
}