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;
int64_t N,X;
int64_t solve(int64_t n,int64_t x){
if(n==0 && x==1)
  return 1;
  else{
  if(x==1)
    return 0;
    else if(2<=x && x<=pow(2,n+1)-2)
      return solve(n-1,x-1);
    else if(x==pow(2,n+1)-1)
      return pow(2,n);
    else if(pow(2,n+1)<=x && x<=pow(2,n+2)-4)
      return pow(2,n)+solve(n-1,x+1-pow(2,n+1));
      else
        return pow(2,n+1)-1;
  }

}
int main(){
  cin>>N>>X;
  int64_t ans=solve(N,X);
  cout<<ans<<endl;
  return 0;
}