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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>

long long int p[51];
long long int a[51];

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

int main(){
  int N;
  long long int X;
  scanf("%d%lld",&N,&X);

  p[0] = 1;
  a[0] = 1;
  for(int i = 1;i <= N;i++){
    p[i] = p[i-1] * 2 + 1;
    a[i] = a[i-1] * 2 + 3;
  }

  printf("%lld\n",f(N,X));

  return 0;
}