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 long long mod = (long long)1e9 + 7;
const long long inf = (long long)1e16;
const long double eps = 1e-12;
const long long N = (int)1e5 + 5;
const long long LOGN = 19;
const long double PI = 3.14159265358979323846;
inline long long mul(long long a, long long b, long long m = mod) {
  return (long long)(a * b) % m;
}
inline long long add(long long a, long long b, long long m = mod) {
  a += b;
  if (a >= m) a -= m;
  if (a < 0) a += m;
  return a;
}
inline long long power(long long a, long long b, long long m = mod) {
  if (b == 0) return 1;
  if (b == 1) return (a % m);
  long long x = power(a, b / 2, m);
  x = mul(x, x, m);
  if (b % 2) x = mul(x, a, m);
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (0) {
    freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\input.txt", "r",
            stdin);
    freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\output.txt", "w",
            stdout);
  }
  long long n;
  cin >> n;
  long long ans = 0ll;
  for (int i = 2; i <= n; i++) {
    for (int j = 2 * i; j <= n; j += i) {
      ans += 4 * j / i;
    }
  }
  cout << ans;
  return 0;
}
