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: Accepted)

import java.util.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {

	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		long x = sc.nextLong();
		
		// 計算
	    long result = dfs(n, x);

		// 出力
		System.out.println(result);
        
	}
	
	public static long pow(long a,long n) {
	    long res=1;
	    while (n>0) {
	        if ((n-n/2*2)==1) {
	            res=(res*a);
	        }
	        a=(a*a);
	        n>>=1;
	    }
	    return res;
	}
	
	public static long len(int n) {
	    return pow(2, n+2) - 3;
	}
	
	public static long p(int n) {
	    return pow(2, n+1) - 1;
	}
	
	public static long dfs(int n, long x) {
	    if(x == 0) return 0;
	    if(n == 0) return 1;
	    long len = len(n-1);
	    if(x <= 1 + len){
	        return dfs(n-1, x-1);
	    }else{
	        return p(n-1) + 1 + dfs(n-1, x-2-len);
	    }
	}
}