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 bool isvowel(char c) {
  c = tolower(c);
  if (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u')
    return 1;
  return 0;
}
const double eps = 0.000001;
const long double pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const long long MOD = 1e18 + 9;
const long long INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 1e6 + 123;
const int M = 22;
const int pri = 997;
const int Magic = 2101;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int n, m, k;
int cnt[N];
string s;
int a[N];
vector<int> g[N];
int u[N];
long long ans, res;
void dfs(int v) {
  u[v] = 1;
  for (auto to : g[v]) {
    if (to > v) {
      res += to / v * 2;
    } else {
      res += v / to * 2;
    }
    if (!u[to]) dfs(to);
  }
}
int main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    for (int j = i + i; j <= n; j += i) {
      g[i].push_back(j);
      g[j].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!u[i]) {
      res = 0;
      dfs(i);
      ans = max(ans, res);
    }
  }
  cout << ans << '\n';
  return 0;
}
