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>
using namespace std;

long long a[51];
long long p[51];

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

}

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

  a[0]=1;
  p[0]=1;
  for(int i=1; i<51; i++) {
    a[i]=2*a[i-1]+3;
    p[i]=2*p[i-1]+1;
  }

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