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;
using System.Collections.Generic;
using System.Linq;
class Program{
	static long[] a,p;
	static void Main(){
		long[] s=Array.ConvertAll(Console.ReadLine().Split(),long.Parse);
		a=new long[s[0]+1];
		p=new long[s[0]+1];
		a[0]=1;p[0]=1;
		for(int i=1;i<=s[0];i++){
			a[i]=a[i-1]*2+3;
			p[i]=p[i-1]*2+1;
		}
		Console.WriteLine(fu(s[0],s[1]));
	}
	static long fu(long n,long x){
		if(x==1){return (n==0)?1:0;}
		else if(1<x&&x<=1+a[n-1]){return fu(n-1,x-1);}
		else if(x==2+a[n-1]){return p[n-1]+1;}
		else if(2+a[n-1]<x&&x<=2+2*a[n-1]){return p[n-1]+1+fu(n-1,x-2-a[n-1]);}
		else {return 2*p[n-1]+1;}
	}
}