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;

ll P[52], S[52];
int main(){
  ll n, k; cin >> n >> k;
  P[0] = S[0] = 1;
  for(ll i = 1; i <= n; i++){
    P[i] = 2*P[i-1] + 1;
    S[i] = 2*S[i-1] + 3;
  }
  ll ans = 0, cnt = n, num = k;
  while(num){
    if(num*2 > S[cnt]){
      if(cnt >= 1) ans += P[cnt-1];
      ans++;
      if(cnt >= 1) num -= S[cnt-1] + 2;
      else break;
    }
    else{
      num--;
    }
    cnt--;
  }
  cout << ans << endl;
}
    