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 int mod = 1000000007;
const long long int N = 200005;
const long long int inf = 1e18;
vector<long long int> vis(N);
vector<vector<long long int>> adj(N);
bool isPrime(long long n) {
  if (n < 2) return false;
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res % mod;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long max(long long a, long long b) {
  long long ans = a > b ? a : b;
  return ans;
}
long long min(long long a, long long b) {
  long long ans = a < b ? a : b;
  return ans;
}
void fun() {}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  fun();
  long long int tt = 1;
  while (tt--) {
    long long int n;
    cin >> n;
    long long int s = 0;
    for (long long int i = 2; i <= n / 2; i++) {
      long long int c = n / i - 1;
      s += (c * i);
    }
    cout << s * 4;
  }
  return 0;
}
