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;
long long int m,s;
long long int t[55][2];

void check(int n){
	if(m==0)return;
	if(m>=t[n][0]){
		m-=t[n][0],s+=t[n][1];
		return;
	}
	m--;
	if(n==0){
		s++;
		return;
	}
	check(n-1);
	if(m==0)return;
	m--,s++;
	check(n-1);
	if(m==0)return;
	m--;
	return;
}	

int main(){
	int n;
	cin>>n>>m;
	t[0][0]=1,t[0][1]=1;
	for(int i=1;i<=n;i++){
		t[i][0]=t[i-1][0]*2+3;
		t[i][1]=t[i-1][1]*2+1;
	}
	check(n);
	cout<<s<<endl;
	return 0;
}
