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>
#include <algorithm>
using namespace std;
long long eat_burger(int n, long long x)
{
  if(n==0){
    if(x==0)
      return 0;
    else
      return 1;
  }
  long long c=((long long)1<<(n+1))-1;
  long long p=(long long)1<<n;
  if(x<=1){
    return 0;
  }else if(x<c){
    return eat_burger(n-1,x-1);
  }else{
    return eat_burger(n-1,x-c)+p;
  }
}
int main()
{
  int n,i;
  long long x,res;
  cin >> n >> x;
  res=eat_burger(n,x);
  cout << res << endl;
  return 0;
}