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;

using ll = long long;

constexpr ll inf = 1<<30;
vector<ll> a(55);
vector<ll> p(55);
ll f(ll n,ll x){
    if(n<=0) return (x<=0)?0:1;
    
    if(x <= a[n-1] + 1){
        return f(n-1,x-1);
    }else{
        return p[n-1] + 1 + f(n-1,x-2-a[n-1]);
    }
    
}
int main(){
    int n;
    ll x;
    cin>>n>>x;
    a[0] = 1;
    p[0] = 1;
    for(int i=0;i<n;++i){
       a[i+1] = 2*a[i]+3;
       p[i+1] = 2*p[i]+1;
    }
    cout << f(n,x) << endl;
    return 0;
}
