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;
int main()
{
	long long a,b;
	cin>>a>>b;
	long long p[55],q[55];
	p[0]=1;
	q[0]=1;
	for(int i=1;i<=50;i++){
		p[i]=p[i-1]*2+3;
		q[i]=q[i-1]*2+1;
	}
	long long ans=0;
	while (b>0){
        if(b==p[a]){
            ans+=q[a];
            b=0;
        }
		else if(b>=p[a]/2+1){
            ans+=q[a-1]+1;
            b-=p[a]/2+1;
        }
		else
        b--;
        a--;
    }
	cout<<ans<<endl;
}