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>
#define REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
vector<ll> b(51), p(51);

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

int main() {
	int n; cin >> n;
	ll x; cin >> x;

	b[0] = p[0] = 1;
	REP(i, 50) {
		p[i+1] = p[i] * 2 + 1;
		b[i+1] = b[i] * 2 + 3;
	}
	cout << f(n, x) << '\n';
    return 0;
}
