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 <cstdio>
using namespace std;
long A[50], P[50];

long f(int N, long X) {
  if (N == 0) {
    if (X <= 0) return 0;
    else return 1;
  } else if (X <= 1 + A[N-1]) {
    return f(N-1, X-1);
  } else {
    return P[N-1] + 1 + f(N-1, X-2-A[N-1]);
  }
}

int main() {
  int N;
  long X;
  scanf("%d %ld", &N, &X);
  A[0] = P[0] = 1;
  for (int i=1; i<N; i++) {
    A[i] = 2 * A[i-1] + 3;
    P[i] = 2 * P[i-1] + 1;
  }
  printf("%ld\n", f(N, X));
}
