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;
const int kN = 100000 + 10;
vector<int> g[kN];
bool used[kN];
void Precalc(int n) {
  for (int i = 2; i < n; ++i) {
    for (int j = 2 * i; j < n; j += i) {
      g[i].push_back(j);
      g[j].push_back(i);
    }
  }
}
void Dfs(int u, vector<int>& comp) {
  if (used[u]) {
    return;
  }
  comp.push_back(u);
  used[u] = true;
  for (int v : g[u]) {
    Dfs(v, comp);
  }
}
void Solve() {
  int n;
  cin >> n;
  Precalc(n + 1);
  long long res = 0;
  vector<int> comp;
  for (int u = 2; u <= n; ++u) {
    if (used[u]) {
      continue;
    }
    comp.clear();
    Dfs(u, comp);
    long long curr = 0;
    for (int i : comp) {
      for (int j : g[i]) {
        if (i > j) {
          continue;
        }
        int uu = min(i, j), vv = max(i, j);
        curr += vv / uu;
      }
    }
    res = max(res, curr);
  }
  cout << res * 4 << '\n';
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests_count = 1;
  for (int test_index = 0; test_index < tests_count; ++test_index) {
    Solve();
  }
  return 0;
}
