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.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long x = sc.nextLong();
        long[] all = new long[n+1];
        long[] pate = new long[n+1];
        all[0] = 1;
        pate[0] = 1;
        for(int i = 1; i <= n; i++) {
            all[i] = all[i-1] * 2 + 3;
            pate[i] = pate[i-1] * 2 + 1;
        }
        System.out.println(dfs(n, x, all, pate));
    }
    private static long dfs(int n, long x, long[] all, long[] pate) {
        if(n == 0) return 1;
        long pateCount = 0;
        // bans
        if(x > 0) x--;
        // level n-1
        if(x > 0) {
            if(x >= all[n-1]) {
                x -= all[n-1];
                pateCount += pate[n-1];
            } else {
                pateCount += dfs(n-1, x, all, pate);
                x = 0;
            }
        }
        // pate
        if(x > 0) {
            x--;
            pateCount++;
        }
        // level n-1
        if(x > 0) {
            if(x >= all[n-1]) {
                x -= all[n-1];
                pateCount += pate[n-1];
            } else {
                pateCount += dfs(n-1, x, all, pate);
                x = 0;
            }
        }
        // bans
        if(x > 0) x--;
        return pateCount;
    }
}
