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 a[55],p[55];
ll ans(ll i,ll nowx){
	if(nowx<=1){
		if(i==0)return 1;
		else return 0;
	}
	if(nowx<=a[i-1]+1)return ans(i-1,nowx-1);
	if(nowx==a[i-1]+2)return p[i-1]+1;
	if(nowx<=2*a[i-1]+2)return p[i-1]+1+ans(i-1,nowx-a[i-1]-2);
	return p[i];
}
int main(){
	ll n,x;
	cin>>n>>x;
	a[0]=p[0]=1;
	for(int i=1;i<=n;i++){
		a[i]=2*a[i-1]+3;
		p[i]=2*p[i-1]+1;
	}
	cout<<ans(n,x);
	return 0;
}
