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 <iostream>
#include <vector>
using namespace std;

long long layer(int x);
long long p_num(int x);
long long solve(int n, long long x);

vector<long long> lvx_pbnum;
vector<long long> lvx_pnum;

int main() {
  long long n, x;
  cin >> n >> x;
  layer(n);
  p_num(n);
  long long result = 0;
/*
  while (n > 0) {
    if (lvx_pbnum[n] < x) {
      result += lvx_pnum[n];
    }
  }
*/


  cout << solve(n, x) << endl;

/*
  int i;
  for (i = 0; i < lvx_pnum.size() ; i++) {
    cout << i << ' ' << lvx_pbnum[i] << ' ' <<  lvx_pnum[i] << endl;
  }
//*/

  return 0;
}

long long layer(int n) {
  if (n == 0) {
    lvx_pbnum.push_back(1);
  } else {
    lvx_pbnum.push_back(layer(n - 1) * 2 + 3);
  }
  return lvx_pbnum[lvx_pbnum.size() - 1];
}

long long p_num(int n) {
  if (n == 0) {
    lvx_pnum.push_back(1);
  } else {
    lvx_pnum.push_back(p_num(n - 1) * 2 + 1);
  }
  return lvx_pnum[lvx_pnum.size() - 1];
}

long long solve(int n, long long x) {
  long long result = 0;
  if (x == 0) {
    return 0;
  } else if (n == 1 && x <= 5) {
    return (x == 5 ? x - 2 : x - 1);
  } else if (n == 0) {
    return 1;
  } else if (lvx_pbnum[n] <= x) {
    result += lvx_pnum[n];
    result += solve(n - 1, x - lvx_pbnum[n]);
  } else {
    result += solve(n - 1, x);
  }
  return result;
}