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
#include <iostream>
using namespace std;

long long a[55], p[55];

long long burger(int n, long long x) {
    if (n == 0) return x <= 0 ? 0 : 1;
    if (x <= 1 + a[n - 1]) return burger(n - 1, x - 1);
    else return p[n - 1] + 1 + burger(n - 1, x - 2 - a[n - 1]);
}

int main() {
    int n;
    long long x;
    cin >> n >> x;

    a[0] = 1, p[0] = 1;
    for (int i = 1; i <= n; i++) {
        a[i] = a[i - 1] * 2 + 3;
        p[i] = p[i - 1] * 2 + 1;
    }

    cout << burger(n, x) << endl;
    return 0;
}