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;
using ll = long long;
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define rep2(i,m,n) for (int i = m; i < (n); ++i)
#define all(x) (x).begin(),(x).end()
template<class T> void chmin(T &a, const T &b) noexcept { if (b < a) a = b;}
template<class T> void chmax(T &a, const T &b) noexcept { if (a < b) a = b;}
template<class T> void drop(const T &x) { std::cout<<x<<endl; exit(0);}
void debug_out() { std::cout << "\n";}
template<class T, class... Args> void debug_out(const T &x, const Args &... args) { std::cout<<x<< " "; debug_out(args...);}
#ifdef _DEBUG
  #define debug(...) debug_out(__VA_ARGS__)
#else
  #define debug(...) 
#endif

struct InitIO{
  InitIO() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << std::fixed << std::setprecision(15);
  }
}init_io;

int main() {
  int n;
  cin >> n;
  vector<ll> cnt(n+1);
  rep2(i, 1, n+1) {
    for (int j = i; j <= n; j += i) cnt[j]++;
  }
  ll ans = 0;
  rep2(i, 1, n+1) {
    ans += cnt[i] * i;
  }
  cout << ans << endl;
  return 0;
}