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;

int hight(int a){return (int)pow(2,a+2)-3;}
int p(int a){return (int)pow(2,a+1)-1;}

int S(int l,int x){//レベルlのバーガーの下からx枚
int ans=0;
if(l==0){return x>=1;}
if(x<=1)return 0;
x--;
if(x<hight(l-1))return S(l-1,x);
x-=hight(l-1);ans+=p(l-1);  
if(x)ans++;
x--;
if(x<=0)return ans;
if(x<hight(l-1))return ans+S(l-1,x);
x-=hight(l-1);ans+=p(l-1);  
return ans;
}
signed main(){
int l,x;cin>>l>>x;
cout<<S(l,x)<<endl;
}