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.io.IOException;

import java.io.BufferedReader;
import java.io.InputStreamReader;

import java.io.PrintWriter;

import java.util.Arrays;

@SuppressWarnings("unchecked")
public class Main {
    static long[] B;
    static long[] P;

    public static void main(String[] args) throws IOException {
        final String s;

        try (
            BufferedReader reader = new BufferedReader(
                new InputStreamReader(System.in))) {
            s = reader.readLine();
        }

        PrintWriter out = new PrintWriter(System.out);

        final String[] sl = s.split(" ");
        int N = Integer.parseInt(sl[0]);
        long X = Long.parseLong(sl[1]);

        B = new long[N];
        P = new long[N];

        B[0] = 1L;
        P[0] = 1L;

        for (int i = 1; i < N; i++) {
            B[i] = 2 * B[i - 1] + 3;
            P[i] = 2 * P[i - 1] + 1;
        }

        long ans = f(N, X);

        out.println(ans);

        out.flush();
    }

    // レベル N
    // 下から X
    // パティ 数
    static long f(int N, long X) {
        if (X <= 0) return 0L;
        if (N == 0) return 1L;
        if (X == 1) return 0L;
        if (1 < X && X <= B[N - 1] + 1) return f(N - 1, X - 1);
        return P[N - 1] + 1 + f(N - 1, X - 2 - B[N - 1]);
    }
}
