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>
using namespace std;
typedef long long ll;
ll p[55];
ll f(ll n, ll x){
    if(n == 0) return 1;
    if(x == 1) return 0;
    if(x == p[n+2] - 3) return p[n+1] - 1;
    ll res = 0;
    if(x == p[n+1] - 1){
        return p[n];
    }
    if(x > p[n+1] - 1){
        res += p[n];
        x -= p[n+1] - 1;
    }else x--;
    
    return res + f(n-1, x);
}
int main() {
    ll n, x;
    cin >> n >> x;
    p[0] = 1;
    for(int i=1;i<55;i++) p[i] = p[i-1] * 2;
    cout << f(n, x) << endl;
	return 0;
}