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;
#define rep(i, n) REP(i, 0, n)
#define REP(i, a, n) for(int i=(a); i<(n); i++)

int N;
ll X;

ll dfs(ll x, int n) {
    if(x < n) return 0;
    if(n == 0) return 1;
    ll m = pow(2, n+2)-3;
    if(x > m-n) return pow(2, n+1) - 1;
    if(x < m/2) return dfs(x-1, n-1);
    else if(x > m/2) return pow(2, n) + dfs(x-m/2-1, n-1);
    else return pow(2, n);
}

int main() {
    cin >> N >> X;
    cout << dfs(X-1, N) << endl;
}
