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 <iostream>
#include <algorithm>
#include <string>
using namespace std;
typedef long long int ll;
ll s[51];
ll p[51];
ll pt(ll a,ll b);
int main(){
  ll n,x;
  cin >> n>>x;
  s[0]=1;//層の数
  p[0]=1;//全体のパティの数
  for(int i=1;i<50;i++){
    s[i]=s[i-1]*2+3;
    p[i]=p[i-1]*2+1;
  }
  cout <<pt(n,x)<<endl;
}
ll pt(ll a,ll b){//levelaのb層目までもパティの数
  if(a==0) {
  if(b<=0) return 0;
  else return 1;
}
  else if(1+s[a-1]>=b) return pt(a-1,b-1);
  else return p[a-1]+1+pt(a-1,b-2-s[a-1]);
}
