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;

#define int long long
signed main(){
    long n,l;
    long ans;
    double x;
    cin >>n>>x;
    l=n;
    ans=0;
    while(x!=0){
        if(x == pow(2,l+2)-3){
            ans+=long(pow(2,l+1))-1;
            break;
        }
        if(x<pow(2,l+1)-1){
            x-=1;
            l-=1;
        }else{
            x-=pow(2,l+1)-1;
            ans+=long(pow(2,l));
            l-=1;

        }
    }
    cout << ans;
    return 0;

  
    
}