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>
#define ll long long
using namespace std;
ll n,m;
ll a[55];
ll dfs(ll x,ll y)
{
	//cout <<x<<" "<<y<<" "<<a[x]<<"\n";
	if (y<=x) return 0;
	else if (x==0) return 1;
	else
	{
	    if (y-1<=a[x-1]) return dfs(x-1,y-1);
	    else if (y-1==a[x-1]+1) return dfs(x-1,y-2)+1;
	    else if (y!=a[x]) return dfs(x-1,a[x-1])+1+dfs(x-1,y-a[x-1]-2);
		else return dfs(x-1,a[x-1])*2+1;
	}
}
int main()
{
	cin >>n>>m;
	a[0]=1;
	for (int i=1;i<=n;i++) a[i]=a[i-1]*2+3;
	cout <<dfs(n,m);
	return 0;
}