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 ll  = long long;
#define int ll
using namespace std;
signed main(){
    vector<int>dp(51,1LL),dp2(51,1LL);
    int n,x; cin >> n >> x;
    for(int i = 1; i < dp.size();++i){
        dp[i] = dp[i-1] + dp[i-1] + 3;
        dp2[i] = dp2[i-1] + dp2[i-1] + 1;
    }
    function<int(int,int)>cal = [&](int i,int x) -> int{
        if(i == 0){ return x > 0;}
        return (x >= dp[i-1] + 2 ? dp2[i-1] + 1 + cal(i-1,x-dp[i-1]-2) : cal(i-1,x-1));
    };
    cout << cal(n,x) << endl;
}


