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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.awt.*;
import java.awt.geom.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        Scanner sc = new Scanner(in);
        int n=sc.nextInt();
        long x=sc.nextLong();
        long[] a=new long[51];
        a[0]=1;
        long[] p=new long[51];
        p[0]=1;
        for (int i = 0; i < 50; i++) {
            a[i+1]=2*a[i]+3;
            p[i+1]=2*p[i]+1;
        }
        out.println(f(a,p,n,x));
    }
    static long f(long[] a,long[] p,int n,long x){
        if(x==1)return 0;
        if(x==3+2*a[n-1])return p[n-1]*2+1;
        if(x==a[n-1]+2)return p[n-1]+1;
        if(1<x&&x<=a[n-1]+1)return f(a,p,n-1,x-1);
        return p[n-1]+1+f(a,p,n-1,x-a[n-1]-2);
    }
}