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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

vector<long long> l(51);
vector<long long> p(51);

long long solve (int n, long long x) {
  assert(l[n] > x && x >= 0);
  if (n == 1) {
    return x;
  }
  if (x == l[n]) {
    return p[n];
  } else if (x == 0) {
    return 0;
  }
  long long ans = 0;
  if (x > l[n - 1] + 1) {
    ans += p[n - 1] + 1;
    x -= l[n - 1] + 1;
  }
  ans += solve(n - 1, x - 1);
  return ans;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n;
  long long x;
  cin >> n >> x;
  l[0] = 1LL;
  p[0] = 1LL;
  for (int i = 1; i <= n; i++) {
    l[i] = l[i - 1] * 2 + 3;
    p[i] = p[i - 1] * 2 + 1;
  }
  cout << solve(n, x) << '\n';
}