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 <cmath>
using namespace std;
using ll = long long;

ll f(ll N, ll X){
        if(X==0) return 0;
        if(X>pow(2,N+2)-3) return pow(2,N+1)-1;
        if(N==0&&X==1) return 1;
        if(X<pow(2,N+1)-1) return f(N-1, X-1);
        if(X==pow(2,N+1)-1) return pow(2,N);
        if(X>pow(2,N+1)-1) return pow(2,N)+f(N-1,X-pow(2,N+1)+1);
}

int main(){
        ll N,X;
        cin >> N >> X;
        cout << f(N,X) << endl;
        return 0;
}