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 <algorithm>

long long b[60], h[60], n, r, ans;

void solve(int now)
{
	if (r <= 0) return;
	if (!now) { --r, ++ans; return; }
	if (r >= h[now]) { r -= h[now], ans += b[now]; return; } 
	--r;
	if (r > 0) solve(now - 1);
	if (r > 0) ++ans, --r;
	if (r > 0) solve(now - 1);
	--r;
}

int main()
{
	scanf("%lld%lld", &n, &r);
	h[0] = 1, b[0] = 1;
	for (int i = 1; i <= 50; ++i) h[i] = (h[i - 1] << 1) + 3, b[i] = 1 + (b[i - 1] << 1);
	solve(n); 
	std::cout << ans;
	return 0;
}