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>
#define int long long
using namespace std;
signed main(){
  int N,X;
  cin>>N>>X;
  vector<int> A(50),B(50);
  A[0]=1;B[0]=1;
  for(int i=1;i<50;i++)
    A[i]=A[i-1]*2+3,B[i]=B[i-1]*2+1;
  int ans=0;
  for(int i=N-1;i>=0;i--){
    if(X>A[i]+1)
      ans++;
    ans+=X/(A[i]+1)*B[i];
    X=X%(A[i]+1)-1;
    if(X<=0)
      break;
  }
  cout<<ans<<endl;
}