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 <cstdio>
using namespace std;
int n;
long long sum=1,qwq=1,m,ans=0;
long long qwqqwq()
{
	for (;n;--n,sum=(sum-3)/2,qwq=(qwq-1)/2)
	{
		if (m==sum/2+1)	return ans+(qwq-1)/2+1;
		else if (m<=sum/2)
			if (m==1) return ans;
			else m--;
		else if (m==sum) return ans+qwq;
		else m=m-sum/2-1,ans+=(qwq-1)/2+1;
	}
	return ans+1;
} 
int main()
{
	scanf("%d%lld",&n,&m);
	for (int i=1;i<=n;++i) sum=sum*2+3,qwq=qwq*2+1;
	cout<<qwqqwq();
}