You are given a positive integer n greater or equal to 2. For every pair of integers a and b (2 ≤ |a|, |b| ≤ n), you can transform a into b if and only if there exists an integer x such that 1 < |x| and (a ⋅ x = b or b ⋅ x = a), where |x| denotes the absolute value of x.

After such a transformation, your score increases by |x| points and you are not allowed to transform a into b nor b into a anymore.

Initially, you have a score of 0. You can start at any integer and transform it as many times as you like. What is the maximum score you can achieve?

Input

A single line contains a single integer n (2 ≤ n ≤ 100 000) — the given integer described above.

Output

Print an only integer — the maximum score that can be achieved with the transformations. If it is not possible to perform even a single transformation for all possible starting integers, print 0.

Examples

Input

4


Output

8

Input

6


Output

28

Input

2


Output

0

Note

In the first example, the transformations are 2 → 4 → (-2) → (-4) → 2.

In the third example, it is impossible to perform even a single transformation.
#include <bits/stdc++.h>
using namespace std;
inline pair<long long, long long> mp(long long a, long long b) {
  pair<long long, long long> temp;
  temp.first = a;
  temp.second = b;
  return temp;
}
long long read_int() {
  char r;
  bool start = false, neg = false;
  long long ret = 0;
  while (true) {
    r = getchar();
    if ((r - '0' < 0 || r - '0' > 9) && r != '-' && !start) {
      continue;
    }
    if ((r - '0' < 0 || r - '0' > 9) && r != '-' && start) {
      break;
    }
    if (start) ret *= 10;
    start = true;
    if (r == '-')
      neg = true;
    else
      ret += r - '0';
  }
  if (!neg)
    return ret;
  else
    return -ret;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long ans = 0;
  for (long long i = 2; i <= n / 2; i++) {
    ans += 4 * (n / i - 1) * i;
  }
  cout << ans;
}
