#include <iostream>
#include <bits/stdc++.h>
using namespace std;
#define int         long long   
#define double      long double
#define endl        "\n"
#define pb          push_back
#define PI          3.1415926535897932384626433832795l
#define F           first
#define S           second
#define mp          make_pair
#define f(i,n)      for(int i=0;i<n;i++)
#define fastio      ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define all(v)      (v).begin(),(v).end()
#define rall(v)     (v).rbegin(),(v).rend()
#define gcd(a,b)    __gcd((a),(b))
#define minn(v)     *min_element(v.begin(), v.end());
#define maxx(v)     *max_element(v.begin(), v.end());
#define print(x)    cout<<(x)<<endl;
#define sum(v)+x    accumulate(v.begin(), v.end(),x);
typedef pair<int,int> pii;  
typedef vector<int> vi;
//x.substr(start, length);
/*
Create a vector containing n 
vectors of size m.  
---vector<vector<int> > vec( n , vector<int> (m, 0));  
*/
 
/*left shifting an integer “x” with an integer “y” (x<<y) 
    is equivalent to multiplying x with 2^y (2 raise to power y).*/
/*Similarly right shifting (x>>y) is equivalent 
    to dividing x with 2^y.
*/
 
struct type{
    int k, i, j;
};
//to check if a no is prime or not?
bool isPrime(int n) 
{ 
    if (n <= 1) 
        return false; 
    if (n <= 3) 
        return true; 
    if (n % 2 == 0 || n % 3 == 0) 
        return false; 
  
    for (int i = 5; i * i <= n; i = i + 6) 
        if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
  
    return true; 
}
 
// minimum subaaray sum of size k-sliding window approach
int findSubarray(int arr[], int n, int k)
{
	int window_sum = 0;
	int min_window = INT_MAX;
	int last = 0;
	for (int i = 0; i < n; i++)
	{
		window_sum += arr[i];
		if (i + 1 >= k)
		{
			if (min_window > window_sum)
			{
				min_window = window_sum;
				last = i;
			}
			window_sum -= arr[i + 1 - k];
		}
	}
	int sum=0;
	for(int i=last-k+1;i<=last;i++)
	{
	    sum+=arr[i];
	}
	return sum;
}
//finds the next permutation to given sequence of nos
int nextpermutation(vi v)
{
    vi t=v;
    sort(all(t));
    int x=0;
    while(true)
    {
        x++;
        if(t==v)
        {
            return x;
        }
        next_permutation(all(t));
    }
}
vector<int>factors(int n) 
{ 
    // Note that this loop runs till square root 
    vi ans;
    for (int i=1; i<=sqrt(n); i++) 
    { 
        if (n%i == 0) 
        { 
            if (n == i*i )ans.pb(i);
            else
            {
                ans.pb(i);
                if (i!=n/i)ans.pb(n/i);
            }
        } 
    } 
    return ans;
}
int power(int x, int y, int p)  
{  
    int res = 1;     // Initialize result  
  
    x = x % p; // Update x if it is more than or  
                // equal to p 
   
    if (x == 0) return 0; // In case x is divisible by p; 
  
    while (y > 0)  
    {  
        // If y is odd, multiply x with result  
        if (y & 1)  
            res = (res*x) % p;  
  
        // y must be even now  
        y = y/2; // y = y/2  
        x = (x*x) % p;  
    }  
    return res;  
}
// Recursive C++ program to check if a string is subsequence of another string 
bool isSubSequence(char str1[], char str2[], int m, int n) 
{ 
    // Base Cases 
    if (m == 0) return true; 
    if (n == 0) return false; 
  
    // If last characters of two strings are matching 
    if (str1[m-1] == str2[n-1]) 
        return isSubSequence(str1, str2, m-1, n-1); 
  
    // If last characters are not matching 
    return isSubSequence(str1, str2, m, n-1); 
} 
void SieveOfEratosthenes(int n, bool prime[], 
                         bool primesquare[], int a[]) 
{ 
    // Create a boolean array "prime[0..n]" and 
    // initialize all entries it as true. A value 
    // in prime[i] will finally be false if i is 
    // Not a prime, else true. 
    for (int i = 2; i <= n; i++) 
        prime[i] = true; 
  
    // Create a boolean array "primesquare[0..n*n+1]" 
    // and initialize all entries it as false. A value 
    // in squareprime[i] will finally be true if i is 
    // square of prime, else false. 
    for (int i = 0; i <= (n * n + 1); i++) 
        primesquare[i] = false; 
  
    // 1 is not a prime number 
    prime[1] = false; 
  
    for (int p = 2; p * p <= n; p++) { 
        // If prime[p] is not changed, then 
        // it is a prime 
        if (prime[p] == true) { 
            // Update all multiples of p 
            for (int i = p * 2; i <= n; i += p) 
                prime[i] = false; 
        } 
    } 
  
    int j = 0; 
    for (int p = 2; p <= n; p++) { 
        if (prime[p]) { 
            // Storing primes in an array 
            a[j] = p; 
  
            // Update value in primesquare[p*p], 
            // if p is prime. 
            primesquare[p * p] = true; 
            j++; 
        } 
    } 
} 
  
// Function to count divisors 
int countDivisors(int n) 
{ 
    // If number is 1, then it will have only 1 
    // as a factor. So, total factors will be 1. 
    if (n == 1) 
        return 1; 
  
    bool prime[n + 1], primesquare[n * n + 1]; 
  
    int a[n]; // for storing primes upto n 
  
    // Calling SieveOfEratosthenes to store prime 
    // factors of n and to store square of prime 
    // factors of n 
    SieveOfEratosthenes(n, prime, primesquare, a); 
  
    // ans will contain total number of distinct 
    // divisors 
    int ans = 1; 
  
    // Loop for counting factors of n 
    for (int i = 0;; i++) { 
        // a[i] is not less than cube root n 
        if (a[i] * a[i] * a[i] > n) 
            break; 
  
        // Calculating power of a[i] in n. 
        int cnt = 1; // cnt is power of prime a[i] in n. 
        while (n % a[i] == 0) // if a[i] is a factor of n 
        { 
            n = n / a[i]; 
            cnt = cnt + 1; // incrementing power 
        } 
  
        // Calculating number of divisors 
        // If n = a^p * b^q then total divisors of n 
        // are (p+1)*(q+1) 
        ans = ans * cnt; 
    } 
  
    // if a[i] is greater than cube root of n 
  
    // First case 
    if (prime[n]) 
        ans = ans * 2; 
  
    // Second case 
    else if (primesquare[n]) 
        ans = ans * 3; 
  
    // Third casse 
    else if (n != 1) 
        ans = ans * 4; 
  
    return ans; // Total divisors 
} 
void findDivisors(int n) 
{ 
  
    // Array to store the count 
    // of divisors 
    int div[n + 1]; 
    memset(div, 0, sizeof div); 
  
    // For every number from 1 to n 
    for (int i = 1; i <= n; i++) { 
  
        // Increase divisors count for 
        // every number divisible by i 
        for (int j = 1; j * i <= n; j++) 
            div[i * j]++; 
    } 
  
    // Print the divisors 
    int ans=0;
    for (int i = 1; i <= n; i++) 
        ans+= i*div[i];
    cout<<ans<<endl;    
}
signed main() 
{
    fastio;
    cout << fixed << setprecision(12);
    int n;
    cin>>n;
    findDivisors(n);
	return 0;
}

Measured memory_bytes: 81764