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;
int n;
long long x, t[51] = {1}, p[51] = {1};

long long dp(int n, long long x){
    if(n == 0){
        return x > 0 ? 1 : 0;
    }else if(x < t[n - 1] + 2){
        return dp(n - 1, x - 1);
    }else{
        return p[n - 1] + 1 + dp(n - 1, x - t[n - 1] - 2);
    }
}

int main(){
    cin >> n >> x;
    for(int i = 1; i <= n; i++){
        t[i] = t[i - 1] * 2 + 3;
        p[i] = p[i - 1] * 2 + 1;
    }
    cout << dp(n, x) << endl;
}
