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>
#define int long long
using namespace std;

int n,x,a[51],b[51],sum,l;

int dfs(int l,int r,int B){
	if(r<=x)return a[B];
	if(x<l)return 0;
	int m=(l+r)/2;
	return dfs(l+1,m-1,B-1)+dfs(m,m,0)+dfs(m+1,r-1,B-1);
}

signed main(){
	cin>>n>>x;
	x--;
	a[0]=1;
	b[0]=1;
	a[1]=3;
	b[1]=5;
	for(int i=2;i<=50;i++){
		a[i]=a[i-1]*2+1;
		b[i]=b[i-1]*2+3;
	}
	cout<<dfs(0,b[n]-1,n)<<endl;
}
