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 ll;

ll pn[51],bn[51];

ll solve(ll n,ll x){
	if(n==0)return 1;
	else if(x<=1)return 0;
	else if(x<=1+bn[n-1])return solve(n-1,x-1);
	else if(x==bn[n-1]+2)return pn[n-1]+1;
	else if(x<=bn[n]-1)return pn[n-1]+1+solve(n-1,x-2-bn[n-1]);
	else return pn[n];
}

int main(){
	ll n,x;
	cin>>n>>x;
	pn[0]=1;bn[0]=1;
	for(int i=0;i<50;i++){
		pn[i+1]=1+2*pn[i];
		bn[i+1]=3+2*bn[i];
	}

	cout<<solve(n,x)<<endl;
	return 0;
}
