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;
long long a[60],p[60];
long long f(long long n,long long x)
{
    if(n==0)
    {
        if(x<=0)
			return 0;
		return 1;
	}
    else if(x<=1+a[n-1])
		return f(n-1,x-1);
    return p[n-1]+1+f(n-1,x-2-a[n-1]);
}
int main()
{
	int i; 
	long long n,x;
	scanf("%lld %lld",&n,&x);
	a[0]=p[0]=1;
	for(i=1;i<n;i++)
	{
		a[i]=a[i-1]*2+3;
		p[i]=p[i-1]*2+1;
	}
	printf("%lld",f(n,x));
	return 0;
}