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.

Examples

Input

2 7


Output

4


Input

1 1


Output

0


Input

50 4321098765432109


Output

2160549382716056
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);
        }

    }
}

