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>

#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

ll rec(ll n, ll x) {
  ll all = (1LL << n + 1) - 3;
  ll p = (1LL << n) - 1;
  if (x == 0)
    return 0;
  else if (x <= all + 1)
    return rec(n - 1, x - 1);
  else if (x <= all + 2)
    return p + 1;
  else if (x <= 2 * (all + 1))
    return p + 1 + rec(n - 1, x - all - 2);
  else
    return p * 2 + 1;
}

int main() {
  ll n, x;
  cin >> n >> x;

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