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 <iostream>
using namespace std;
typedef long long ll;
ll a[51], p[51];

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