Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

2 7

Sample Output 1

4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).

Sample Input 2

1 1

Sample Output 2

0

The bottom-most layer of a level-1 burger is a bun.

Sample Input 3

50 4321098765432109

Sample Output 3

2160549382716056

A level-50 burger is rather thick, to the extent that the number of its layers does not fit into a 32-bit integer.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void) {

  long n,x;
  scanf("%ld %ld", &n, &x);
  long p[n+1],b[n+1],all[n+1];
  p[0] = 1;
  b[0] = 0;
  all[0] = 1;
  for (int i = 1; i <= n; i++) {
    p[i] = p[i-1]*2+1;
    b[i] = b[i-1]*2+2;
    all[i] = p[i]+b[i];
  }
  long eat = 0;
  for (int i = n; i >= 0; i--) {
    if (x < all[i]) {
      if (x > 0) {
        x--;
      } else {
        break;
      }
    } else {
      eat += p[i];
      x -= all[i];
      if (x > 0) {
        eat++;
        x--;
      }
      if (x >= all[i]) {
        eat += p[i];
        x -= all[i];
      }
      if (x > 0) {
        x--;
      }
      if (x == 0) {
        break;
      }
    }
  }
  printf("%ld\n", eat);

  return 0;
}