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;
using ll = long long;
int main(){
    ll N,K,i;
    cin >> N >> K;
  
    ll A[N + 1];
    ll P[N + 1];
    A[0] = 1;
    P[0] = 1;
    for(i = 1;i <= N;i++){
        A[i] = 2 * A[i - 1] + 3;
        P[i] = 2 * P[i - 1] + 1;
    }
    ll count = 0;
    for(i = N; i > 0; i--){
        if(K == 0) break;
        K--;
        if(K <= A[i - 1]) continue;
        K -=A[i - 1] + 1;
        count += P[i - 1] + 1;
    }
    if(K) count++;
    cout << count << "\n";
    return 0;
}