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 <vector>
using namespace std;
using ll = long long;

vector<ll> lay;
vector<ll> p;
ll n, k;

ll fun(ll n, ll x){
	if (n == 0)return (x <= 0 ? 0 : 1);
	if (x <= lay[n - 1] + 1)return fun(n - 1, x - 1);
	return p[n - 1] + 1 + fun(n - 1, x - 2 - lay[n - 1]);
	
}

int main(){
	cin >> n >> k;
	lay.push_back(1), p.push_back(1);
	for (int i = 1; i <= n; i++){
		ll l = lay[i - 1] * 2 + 3;
		ll pp = p[i - 1] * 2 + 1;
		lay.push_back(l);
		p.push_back(pp);
	}
	cout << fun(n, k) << endl;
}