#include <bits/stdc++.h>
using namespace std;

#define pii  pair<int, int>
#define vi vector<int >
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define sz(x) (x).size()
#define all(x) (x).begin(), (x).end() //Forward traversal
#define rall(x) (x).rbegin, (x).rend() //reverse traversal

#define ll long long
#define ld long double

const ld EPS = 10e-9;
const int INF = 1e9 + 5;
const long long INFLL = (ll)INF * (ll)INF;
const ll MOD = 1e9 + 7;
const ld PI = 3.14159265358979323846;

struct HASH{
    size_t operator()(const pair<int,int>&x)const{
        return hash<long long>()(((long long)x.first)^(((long long)x.second)<<32));
    }
};

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    os << "[";
    for (int i = 0; i < (int)v.size(); ++i) {
        os << v[i];
        if (i != (int)v.size() - 1)
            os << ", ";
    }
    os << "]\n";
    return os;
}

template <typename T>
ostream& operator<<(ostream& os, const set<T>& v)
{
    os << "{";
    for (const auto& it : v) {
        os << it;
        if (it != *v.rbegin())
            os << ", ";
    }
    os << "}\n";
    return os;
}

template <typename T>
ostream& operator<<(ostream& os, const unordered_set<T>& v)
{
    os << "{";
    for (const auto& it : v) {
        os << it;
        if (it != *v.rbegin())
            os << ", ";
    }
    os << "}\n";
    return os;
}

template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v)
{
    os << "{";
    for (const auto& it : v) {
        os << it.first << " : "
           << it.second;

        if( it != *v.rbegin())
            os  << ", ";
    }

    os << "}\n";
    return os;
}

template <typename T, typename S>
ostream& operator<<(ostream& os, const unordered_map<T, S>& v)
{
    os << "{";
    for (const auto& it : v) {
        os << it.first << " : "
           << it.second;

        if( it != *v.rbegin())
            os  << ", ";
    }

    os << "}\n";
    return os;
}

template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v)
{
    os << "(";
    os << v.first << ", "
       << v.second << ")";
    os << "\n";
    return os;
}

#define DEBUG

#ifdef DEBUG
#define debug(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << " : " << arg1;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define debug(...)
#endif


// __gcd(x, y)

// idea behind binary search: there is monotonically increasing predicate and range [l, r]. Find
// x in range such that x is smallest value for which predicate holds true.
//            int lo = 1, hi = 1e9;
//            while(lo < hi)
//            {
//                ll mid = (lo+hi) / 2;
//                if(predicate(mid))
//                {
//                    hi = mid;
//                }
//                else
//                {
//                    lo = mid+1;
//                }
//            }
//            assert(lo == hi);

// OR: given monotonically decreasing predicate, find x in range such that x is largest value for
// which predicate holds true.

//int x = 0;
//for(int s = max_step; s >= 1; s /= 2)
//{
//    while(predicate(x+s)) x += s;
//}
// after this, x is the largest value for which predicate is true

// How to calculate mid element:
// floor((l+r) / 2) might overflow, use l + floor((r-l)/2)



//ll l = 0, r = s.size()-1;
//ll ans = -1;
//while(l < r)
//{
//    ll mid = l + (r-l) / 2;
//    if(P(mid)) {
//        r = mid;
//        ans = mid;
//    }
//    else
//    {
//        l = mid + 1;
//    }
//}
//
// AND
//
//ll l = 0, r = s.size()-1;
//ll ans = -1;
//while(l <= r)
//{
//    ll mid = l + (r-l) / 2;
//    if(P(mid)) {
//        r = mid-1;
//        ans = mid;
//    }
//    else
//    {
//        l = mid + 1;
//    }
//}
//
// ARE EQUIVALENT


ll getNum(ll x) {
    ll cnt = 0;
    for(int i = 1; i <= ceil(pow(x, 0.25)); i++) {
        if(x%i == 0) {
            cnt++;
            if(x/i != i)
                cnt++;
        }
    }

    return cnt;
}

ll smallestDiv(ll x) {
    if(x == 3)
        return 3;
    ll mi = x;
    for(ll i = 1; i <= sqrt(x); i++)
    {
        if(x%i == 0) {
            if(i != 1)
                mi = min(mi, i);
        }
    }

    return mi;
}

#define MAXN   100000001

// stores smallest prime factor for every number
int spf[MAXN];

// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
void sieve()
{
    spf[1] = 1;
    for (int i=2; i<MAXN; i++)

        // marking smallest prime factor for every
        // number to be itself.
        spf[i] = i;

    // separately marking spf for every even
    // number as 2
    for (int i=4; i<MAXN; i+=2)
        spf[i] = 2;

    for (int i=3; i*i<MAXN; i++)
    {
        // checking if i is prime
        if (spf[i] == i)
        {
            // marking SPF for all numbers divisible by i
            for (int j=i*i; j<MAXN; j+=i)

                // marking spf[j] if it is not
                // previously marked
                if (spf[j]==j)
                    spf[j] = i;
        }
    }
}

// A O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
vector<int> getFactorization(int x)
{
    vector<int> ret;
    while (x != 1)
    {
        ret.push_back(spf[x]);
        x = x / spf[x];
    }
    return ret;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

//    sieve();
    ll n;
    cin >> n;


    ll ans = 1;
    vector<ll> divs(n+1, 0);
    divs[1] = 1;
    for(int i = 2; i <= n; i++) {
        for(int j = i; j <= n; j+=i) {
            divs[j]++;
        }
    }

    for(int i = 2; i <= n; i++) {
        ans += (divs[i]+1)*i;
    }
//    for(ll i = 2; i <= n; i++) {

//         vector <int> p = getFactorization(i);
//         bool same = true;
//         for(int j = 1; j < p.size(); j++) {
//            if(p[0] != p[j]) {
//                same = false;
//                break;
//            }
//         }
//         if(same) {
//             int k = p.size();
//             if(k%2 == 0) {
//                 ans += i * ((k/2+1)*2+3);
//             }
//             else {
//                 ans += i* ((k/2 * 2) + 2);
//             }
//         }
//         else {
//             ans += i*(p.size()*2+2);
//         }
//    }

    cout << ans << endl;


    return 0;
}


Measured memory_bytes: 81344