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;
using ll=long long;

const int MAXN=51;
int n; ll k;
ll p[MAXN],sum[MAXN];

ll calc(int n,ll k){
  if(n==0){
    if(k<=0) return 0;
    else return 1;
  }else if(k<=sum[n-1]+1){
    return calc(n-1,k-1);
  }else{
    return p[n-1]+1+calc(n-1,k-sum[n-1]-2);
  }
}

int main(){
  cin>>n>>k;
  sum[0]=1;
  p[0]=1;
  for(ll i=1;i<=n;i++){
    sum[i]=2*sum[i-1]+3;
    p[i]=p[i-1]*2+1;
  }
  cout<<calc(n,k)<<endl;
}