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 ll long long
using namespace std;

int main(){
    ll n, x;
    cin >> n >> x;
    vector<ll> p(n + 1, 1LL), b(n + 1, 1LL);
    for(int i = 1; i <= n; i++){
        p[i] = p[i-1] + 1 + p[i-1];
        b[i] = 1 + b[i-1] + 1 + b[i-1] + 1;
    }
    ll sum = 0LL;
    for(int i = n; i >= 1; i--){
        if(x == 0) break;
        x--;
        if(x <= b[i-1]) continue;
        sum += p[i-1] + 1;
        x -= b[i-1] + 1;
    }
    if(x) sum++;
    cout << sum << endl;
    return 0;
}
