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 n,m,ans,a[55],b[55];
int main(){
    ios::sync_with_stdio(false);
    cin >> n >> m;
    a[0]=b[0]=1;
    for(int i=1;i<=n;i++)
    	a[i]=a[i-1]*2+3,b[i]=b[i-1]*2+1;
    for(int i=n;i>=0;i--){
    	if(m<=0)break;
    	if(m==a[i]){ans+=b[i];break;}
    	if(m==1)break;
    	if(m<=a[i-1]+1)m--;
    	else ans+=b[i-1]+1,m-=a[i-1]+2;
	}
    cout << ans;
    return 0;
}