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: Wrong Answer)

#include <stdio.h>

int main(){
  int n, i, j;
  long long int x, left_s=0, left_p=0, ans=0, s[70], p[70];
  scanf("%d %lld", &n, &x);

  s[0]=1;
  p[0]=1;
  for(i=1; i<=n; i++){
    s[i]=2*s[i-1]+3;
    p[i]=2*p[i-1]+1;
  }

  for(i=n; i>0; i--){
    if(x==left_s+1){
      ans=left_p;
      break;
    }
    else if(left_s+1 < x && x < left_s+1+s[i]/2){
      left_s++;
    }
    else if(x==left_s+1+s[i]/2){
      ans=left_p+1+p[i]/2;
      break;
    }
    else if(left_s+1+s[i]/2 < x && x < left_s+s[i]/2){
      left_s=left_s+1+s[i]/2;
      left_p=left_p+p[i]/2;
    }
    else{
      ans=left_p+p[i]/2;
      break;
    }
  }
  printf("%d\n", ans);
  return 0;
}