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>
#include<vector>
#define ll long long
using namespace std;


ll length[50];
ll p[50];

ll paty(int L,ll X){
    if(X == 0) return 0;
    if(L == 0) return 1;
    if(X <= length[L-1]+1) return paty(L-1,X-1);
    else return p[L-1] + 1 + paty(L-1,X-2-length[L-1]);
}

int main(){
    length[0] = 1;
    p[0] = 1;
    for(int i=1;i<50;i++){
        length[i] = 2*length[i-1]+3;
        p[i] = 2*p[i-1]+1;
    }
    int N;
    ll X;
    cin >> N >> X;
    cout << paty(N,X) << endl;
    return 0;
}
