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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
typedef long long ll;

ll a[51], p[51];
ll d(int n, ll x)
{
  if (n == 0)
    return 1;
  else if (x == 1)
    return 0;
  else if (x <= a[n - 1] + 1)
    return d(n - 1, x - 1);
  else if (x == a[n - 1] + 2)
    return p[n - 1] + 1;
  else if (a[n - 1] + 2 < x && x <= 2 * a[n - 1] + 2)
    return p[n - 1] + 1 + d(n - 1, x - 2 - a[n - 1]);
  else
    return 2 * p[n - 1] + 3;
}

int main()
{
  int N;
  ll X;
  cin >> N >> X;
  a[0] = p[0] = 1;
  rep(i, N)
  {
    a[i + 1] = 2 * a[i] + 3;
    p[i + 1] = 2 * p[i] + 1;
  }
  cout << d(N, X) << endl;
  return 0;
}