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;
const ll MOD=(ll)1e9+7;

ll l[55],p[55];

ll dfs(ll N,ll X){
    if(N==0){
        return ((X<=0)?0:1);
    }else if(X<=1+l[N-1]){
        return dfs(N-1,X-1);
    }else{
        return p[N-1]+1+dfs(N-1,X-l[N-1]-2);
    }
}

int main(){
    ll N,X; cin>>N>>X;
    l[0]=1; p[0]=1;
    for(int i=1;i<=N;++i){
        l[i]=2*l[i-1]+3;
        p[i]=2*p[i-1]+1;
    }

    cout<<dfs(N,X)<<endl;

    return 0;
}

