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 calc(int n,long x,vector<long>p,vector<long>b){
  long ans = 0;
  if(n==0||x==0){
    return min(x,1l);
  }else{
    if(x<=1+p[n-1]+b[n-1])return calc(n-1,x-1,p,b);
    else return p[n-1]+1+calc(n-1,x-2-b[n-1]-p[n-1],p,b);
  }
}

int main(){
  int n;
  long x;
  cin >> n >> x;
  vector<long>p(51);
  vector<long>b(51);
  int i;
  p[0] = 1;
  b[0] = 0;
  for(i=1;i<=50;i++){
    p[i] = 2*p[i-1]+1;
    b[i] = 2*b[i-1]+2;
  }
  cout << calc(n,x,p,b) << endl;
}
