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;
typedef long long int ll;
vector<ll> p(51),al(51);


int main(){
	ll n,x;
	cin>>n>>x;
	p[0] =al[0] = 1;
	for(int  i= 1;i<n+1;i++){
		p[i] = p[i-1]*2+1;
		al[i] = al[i-1]*2+3;
		//cout<<p[i]<<' '<<al[i]<<endl;
	}
	ll ans = 0;

	while(x>0&&n>=0){
		if(x>=al[n]){
		//cout<<n<<endl;
		x-=al[n];
		ans+=p[n];
			if(x>0){
				x--;
				ans++;
			}
		}else{
		x--;
		n--;
		}
	}
	cout<<ans<<endl;
	return 0;
}
