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;
typedef long long ll;
vector<ll> a(51,1),p(51,1);

ll f(ll n,ll x)
{
  if(x==0) return 0;
  else if(x==a[n]) return p[n];
  else if(x<=a[n-1]+1)
  {
    return f(n-1,x-1);
  }
  else if(x==a[n-1]+2)
  {
    return p[n-1]+1;
  }
  else
  {
    return p[n-1]+1+f(n-1,x-a[n-1]-2);
  }
}

int main()
{
  ll n,x;
  cin>>n>>x;
  for (int i = 1; i < 51; ++i)
  {
    a[i]=2*a[i-1]+3;
    p[i]=2*p[i-1]+1;
  }
  cout<<f(n,x)<<endl;
}