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;
const int maxn = 60;
long long len[maxn], ps[maxn];
long long f(long long n, long long k){
	if(n == 0) return (k <= 0 ? 0 : 1);
	else if(k <= 1 + len[n - 1]) return f(n-1,k-1);
	return ps[n - 1] + 1 + f(n - 1, k - 2 - len[n - 1]);
}
int main() {
	len[0] = 1, ps[0] = 1;
	for(int i = 1; i < maxn; ++i){
		len[i] = 1 + len[i - 1] + 1 + len[i - 1] + 1;
		ps[i] = ps[i - 1] + 1 + ps[i - 1];
	}
	long long n, k;
	cin >> n >> k;
	cout << f(n, k);
	return 0;
}