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;
long long int prime[1000001], a[1000001];
vector<long long int> ar;
void sieve() {
  memset(prime, 1, sizeof(prime));
  prime[0] = prime[1] = 0;
  long long int i, j;
  for (i = 2; i <= sqrt(1000010); i++) {
    if (prime[i]) {
      for (j = i * 2; j < 1000010; j += i) {
        prime[j] = 0;
      }
    }
  }
  for (i = 0; i < 1000010; i++) {
    if (prime[i]) {
      ar.push_back(i);
    }
  }
}
int main() {
  sieve();
  long long int n, i, j, ans, temp;
  cin >> n;
  for (i = 2; i <= n; i++) {
    if (prime[i]) {
      a[i] = 0;
    } else {
      ans = 0;
      for (j = 2; j <= sqrt(i); j++) {
        if (i % j == 0) {
          ans += (i / j) * 4;
          temp = i / j;
          if (i != temp * temp) {
            ans += (i / temp) * 4;
          }
        }
      }
      a[i] = ans;
    }
  }
  ans = 0;
  for (i = 2; i <= n; i++) {
    ans += a[i];
  }
  cout << ans;
  return 0;
}
