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>
#define N 55
using namespace std;

typedef long long ll;
ll n, x, d1[N]{1}, d2[N]{1};

ll f(ll p) {
	ll re = 0;
	if (x <= 0) return 0;
	if (d1[p] <= x) {x -= d1[p]; return d2[p];}
	x--;
	re += f(p - 1) + f(0) + f(p - 1);
	x--;
	return re;
}

int main()
{
	ll i;
	cin >> n >> x;
	for (i = 1; i <= n; i++) {
		d1[i] = d1[i - 1] * 2 + 3;
		d2[i] = d2[i - 1] * 2 + 1;
	}
    cout << f(n) << endl;
    return 0;
}