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 runrun(long long a,long long b){
  if(a==0){return 1;}
  if(b<=1){return 0;}
  long long mon=1,mmon;
  for(int i=0;i<a;++i){
    mon*=2,mon+=3;
  }
  mmon=(mon-3)/2;
  if(mmon+1>=b){return runrun(a-1,b-1);}
  return ((long long)1<<a)+(b==2+mmon ? 0 : runrun(a-1,b-2-mmon));
}

int main(){
  long long a,b;
  cin>>a>>b;
  cout<<runrun(a,b)<<endl;
}
