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>
using namespace std;
typedef long long ll;
ll n, x, s, a[55]={1}, b[55]={1};

void f(ll p, ll q) {
	if(p==0) {s++; return;}
	if(q==1) return;
	ll md = (a[p]+1) / 2;
	if(q < md) f(p-1, q-1);
	else {
		s += b[p-1] + 1;
		if(q>md) f(p-1, q-md);
	}
}

int main() {
	cin>>n>>x;
	for(int i=1; i<=n; i++) {
		a[i] = a[i-1]*2 + 3;
		b[i] = b[i-1]*2 + 1;
	}
	f(n, x);
	cout<<s;
	return 0;
}