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 <bits/stdc++.h>
using namespace std;
using ll = long long;

int main() {
  ll n, x;
  cin >> n >> x;
  vector<ll> s(n + 1), p(n + 1);
  s[0] = p[0] = 1;
  for (int i = 1; i <= n; i++) {
    s[i] = 2 * s[i - 1] + 3;
    p[i] = 2 * p[i - 1] + 1;
  }
  ll r = 0;
  while (x > 0) {
    if (x >= s[n]) {
      r += p[n];
      x -= s[n];
      if (x > 0) {
        r++;
        x--;
      }
    } else if (x < s[n]) {
      n--;
      x--;
    }
  }
  cout << r << endl;
}
