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>
using namespace std;

long long a[51],p[51];

long long f(long long N,long long X){
    if (N == 0)
        if (X <= 0) return 0;
        else return 1;
    else if (X <= 1 + a[N-1])
        return f(N-1, X-1);
    else 
        return p[N-1] + 1 + f(N-1, X-2-a[N-1]);
}

int main(){
    long long N,X;
    cin >> N >> X;

    a[0] = 1;
    p[0] = 1;
    for(int i=1;i<N;i++){
        a[i] = a[i-1] * 2 + 3;
        p[i] = p[i-1] * 2 + 1;
    }

    cout << f(N,X) << endl;
}