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 llint long long int
using namespace std;
llint n,k;
llint len[51],full[51];

llint f(llint n, llint k)
{
	if(k<=0)return 0;
	if(n<=0)return 1;
llint ret=0;
	if(k>len[n]/2)
 return full[n-1]+1+f(n-1,k-len[n]/2-1);
 else return f(n-1,k-1); 
}

int main(void)
{
	len[0]=1;
	for(int i=1;i<51;i++) len[i]=len[i-1]*2+3;
	full[0]=1;
	for(int i=1;i<51;i++) full[i]=full[i-1]*2+1;
	cin>>n>>k;
	cout<<f(n,k)<<endl;
	return 0;
}