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>
typedef long long ll;
using namespace std;
ll p[55];
ll a;
ll f(ll n, ll x) {
  // cin >> a;
  //cout << n <<"   "<<x << endl;
  if(x <= 0 || n < -1) return 0;

  ll res = 0;
  
  if(x <= p[n+1] - 2ll) {
    res = f(n-1,x-1ll);
  }
  else {
    res = f(n-1,x-p[n+1]+1) + p[n];
  }
  return res;
}
  
int main() {
  ll n;
  ll x;
  cin >> n >> x;
  p[0] = 1;
  for(int i = 1; i <= n+1; i++) {
    p[i] = p[i-1] * 2;
  }

  cout << f(n,x) << endl;
}
 
  
