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 <stdio.h>
#include <vector>
typedef long long ll;
using namespace std;

int main(void) {
  ll i, j, k, n, x, ans = 0;
  scanf("%lld%lld", &n, &x);
  for(i = n + 2; i >= 3; --i) {
    if((1ll << i) <= x + 3) {
     ans += (1ll << (i - 1)) - 1;
     x -= (1ll << i) - 3;
     if(x) x--, ans++;
    }
    if(x) x--;
  }
  if(x > 0) {
    if(x < 4) ans += x;
    else ans += 3;
  }
  printf("%lld", ans);
  return 0;
}