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> a(n + 1, 1), b(n + 1, 1);
  for (ll i = 1; i <= n; i++) {
    a[i] = 1 + a[i - 1] + 1 + a[i - 1] + 1;
    b[i] = b[i - 1] + 1 + b[i - 1];
  }
  ll ans = 0;
  for (ll i = n; i >= 1; i--) {
    if (x == 0) break;
    x--;
    if (x <= a[i - 1]) continue;
    ans += b[i - 1] + 1;
    x -= a[i - 1] + 1;
  }
  if (x) ans++;
  cout << ans << endl;
  return 0;
}
