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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
long long a[51],b[51],k,ans;
int n;
int main()
{
	a[0]=1;b[0]=1;for(int i=1;i<=50;++i) a[i]=a[i-1]*2+3,b[i]=b[i-1]*2+1;
	scanf("%d%lld",&n,&k);
	for(int i=n;i>=0;--i)
	if(a[i]+n-i<=k) {ans+=b[i];k-=a[i]+n-i;n=i;if(k) {++ans;k--;}}
	else if(k-n+i>=a[i]-i) {ans+=b[i];break;}
	printf("%lld",ans);
	return 0;
}