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>
#define int long long
using namespace std;

int z(int n) {
    return (1LL << (n + 1)) - 3;
}
int p(int n) {
    return (1LL << n) - 1;
}

int ans(int n, int x) {
    if(x <= 0) return 0;
    if(n == 1) return 1;
    if((x - 1) < z(n - 1)) return ans(n - 1, x - 1);
    return p(n - 1) + 1 * ((x - 1 - z(n - 1)) > 0) + ans(n - 1, x - 2 - z(n - 1));
}

signed main() {
    ios::sync_with_stdio(0); cin.tie(0);
    int n, x;
    cin >> n >> x;
    ++n;
    cout << ans(n, x) << endl;
}
