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;

long long burg(long long  N,long long X){
  long long length,point;
  if(X<=0) return 0;
  else if(N==0) return 1;
  length = 4*pow(2,N)-3;
  point = length/2 + 1;
  if(point > X) return burg(N-1,X-1);
  else if(point == X) return (point-1)/2+1;
  else return (point-1)/2+1+burg(N-1,X-point);
}

int main(int argc,char* argv[]){
  long long N,X;
  cin >> N >> X;
  cout << burg(N,X) << endl;

  return 0;
}
