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>
#include <vector>
#define ll long long
using namespace std;
ll N, X, a[55], p[55];

ll f(ll N, ll X) {
  if (N == 0)
    return X <= 0 ? 0 : 1;
  else if (X <= a[N - 1] + 1)
    return f(N - 1, X - 1);
  else
    return p[N - 1] + 1 + f(N - 1, X - a[N - 1] - 2);
}

int main() {
  cin >> N >> X;

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

  cout << f(N, X) << endl;
  return 0;
}