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.Scanner;

public class Main {
    static long[] P = new long[51];
    static long[] B = new long[51];

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        long X = sc.nextLong();
        sc.close();
        // レベルiバーガーに含まれるパティの枚数
        P[0] = 1L;
        // レベルiLバーガーの層の総数
        B[0] = 1L;
        for(int i = 1; i <= 50; i++){
            P[i] = P[i - 1] * 2 + 1L;
            B[i] = B[i - 1] * 2 + 3L;
        }

        System.out.println(f(N, X));
    }

    // レベルNバーガーの下からX層に含まれるパティの枚数を返却する
    private static long f(int N, long X){
        if(N == 0){
            // レベル0バーガーの時
            if(X > 0){ return 1L; } else { return 0L;}
        } else if(X <= 1 + B[N-1]){
            // 食べる層数Xがバン＋レベルN-1バーガー以下の時
            return f(N-1, X-1);
        } else {
            // 食べる層数Xがバン＋レベルN-1バーガーより多い時（一つ目のレベルN-1バーガーを食べきってそれ以降のパティとレベルN-1バーガーを食べるとき）
            return P[N-1] + 1L + f(N-1, X-B[N-1]-2L);
        }

    }
}