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;
typedef long long ll;
const int MOD = 1e9+7,MAX = 1e5;

ll pate[52],pan[52];

int main()
{
	ll n,x;
	cin >> n >> x;
	pate[0] = 1;
	pan[0] = 1;
	for(int i=0;i<51;i++)
	{
		pate[i+1] = pate[i]*2+1;
		pan[i+1] = pan[i]*2+3;
	}
	//cout<<pan[2]<<endl;
	ll ans = 0;
	for(int i =n;i>=0;i--)
	{
		if(x>pan[i]/2)
		{
			ans += pate[i]/2+1;
			x-= pan[i]/2+1;
		}
		else
		{
			x--;
		}
	}
	cout<<ans<<endl;
	return 0;
}