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;

typedef long long ll;
ll n, m, s[55];
ll f(ll p, ll q) {
	ll md;
	if (p == -1 || q <= 0) return 0;
	md = s[p + 1] - 1;
	if (q < md) return f(p - 1, q - 1);
	else return f(p - 1, q - md) + s[p];
}
int main()
{
	ll i;
	cin >> n >>m;
	s[0] = 1;
	for (i = 1; i <= 51; i++) s[i] = s[i - 1] * 2;
	cout << f(n, m);
    return 0;
}
