Problem:
Score : 400 points

Problem Statement

For a positive integer X, let f(X) be the number of positive divisors of X.

Given a positive integer N, find \sum_{K=1}^N K\times f(K).

Constraints

1 \leq N \leq 10^7

Input

Input is given from Standard Input in the following format:

N

Output

Print the value \sum_{K=1}^N K\times f(K).

Sample Input 1

4

Sample Output 1

23

We have f(1)=1, f(2)=2, f(3)=2, and f(4)=3, so the answer is 1\times 1 + 2\times 2 + 3\times 2 + 4\times 3 =23.

Sample Input 2

100

Sample Output 2

26879

Sample Input 3

10000000

Sample Output 3

838627288460105

Watch out for overflows.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define pli pair<ll, int>
#define pil pair<int, ll>
#define pll pair<ll, ll>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
#define endl '\n'
const int inf = 1000000007;
const ll INF = 1e18;
// int mod = 998244353;
int mod = 1000000007;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
template<typename T, typename U> T pow_(T a, U b) { return b ? pow_(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }
template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; }
template<typename T> inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T, class T2> inline void add(T &a, T2 b) { a += b % mod; if (a >= mod) a -= mod; }
template<class T> void operator += (vector<T>& v, vector<T> v2) { rep(i, sz(v2)) v.eb(v2[i]); }

void solve();

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12); // debuged
  int T;
  T = 1;
  while (T--) solve();
}

void solve() {
  int n;
  cin >> n;

  ll ans = 0;
  rep3(i, 1, n + 1) {
    // ans に i の倍数が足される
    int l = n / i * i; // 末項
    int n_ = (l - i) / i + 1;
    // cerr << i << " " << l << " " << n_ << endl;
    ans += (i + l) * 1LL * n_ / 2; // 初項 i, 末項 l, 項数 n_ の等差数列の和
  }
  cout << ans << endl;
}