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 javax.lang.model.util.ElementScanner6;

public class Main{
    static long[] sum, p;
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        
        int N = sc.nextInt();
        long X = sc.nextLong();
        sc.close();

        sum = new long[N];
        p = new long[N];
        for(int i = 0; i < N; i++){
            if(i == 0){
                sum[0] = 1;
                p[0] = 1;
            } else {
                sum[i] = 2*sum[i-1]+3;
                p[i] = 2*p[i-1]+1;
            }
        }
        System.out.println(ans(N, X));
    }
    static long ans(int N, long X){
        if(X == 1){
            return 0;
        } else {
            if(1 <= X && X <= 1+sum[N-1])
                return ans(N-1, X-1);
            else if(X == 2+sum[N-1])
                return p[N-1]+1;
            else if(2+sum[N-1]<X && X <= 2+2*sum[N-1])
                return p[N-1]+1+ans(N-1, X-2-sum[N-1]);
            else 
                return 2*p[N-1]+1;
        }
    }
}