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
///created: 29/May/2019 08:59:18

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int n;
ll x, p[51], f[51];

ll calc(int n, ll x) {
	if (n == 0) return x >= 1;
	if (x <= 1 + p[n - 1]) return calc(n - 1, x - 1);
	return f[n - 1] + calc(n - 1, x - 2 - p[n - 1]) + 1;
}

int main() {
	cin >> n >> x;
	p[0] = 1;
	for (int i = 1; i <= 50; ++i)
		p[i] = 3LL + 2LL * p[i - 1];
	f[0] = 1;
	for (int i = 1; i <= 50; ++i)
		f[i] = 1LL + 2LL * f[i - 1];
	printf("%lld", calc(n, x));
	return 0;
}
