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;
typedef long long ll;

vector<ll> a(51), p(51);

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

int main() {
    ll N, X;
    cin >> N >> X;


    a[0] = 1; p[0] = 1;
    for (int i = 1; i < N; i++) {
        a[i] = a[i-1]*2 + 3;
        p[i] = p[i-1]*2 + 1;
    }


    ll res = baga(N, X);

    cout << res << endl;

}
