// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long double ld;

typedef pair<ll,ll> P;

typedef pair<int,int> Pi;

#define rep(i,n) for(ll i=0;i<n;i++)

#define FOR(i,a,b) for(ll i=a;i<b;i++)

#define fi first

#define se second

#define endl "\n"



template<typename T> inline bool chmax(T &a, T b){if(a<b){a=b;return true;}return false;}

template<typename T> inline bool chmin(T &a, T b){if(a>b){a=b;return true;}return false;}

template<typename T> ostream& operator<<(ostream& s,const complex<T>& d) {return s<<"("<<d.real()<<", "<<d.imag()<< ")";}

template<typename T1, typename T2> ostream& operator<<(ostream& s,const pair<T1,T2>& d) {return s<<"("<<d.first<<", "<<d.second<<")";}

template<typename T> ostream& operator<<(ostream& s, const vector<T>& d){int len=d.size();rep(i,len){s<<d[i];if(i<len-1) s<<" ";}return s;}

template<typename T> ostream& operator<<(ostream& s,const vector<vector<T>>& d){int len=d.size();rep(i,len){s<<d[i]<<endl;}return s;}

template<typename T> ostream& operator<<(ostream& s,const set<T>& v){s<<"{ ";for(auto itr=v.begin();itr!=v.end();++itr) {if (itr!=v.begin()) {s<< ", ";}s<<(*itr);}s<<" }";return s;}

template<typename T> ostream& operator<<(ostream& s,const multiset<T>& v){s<<"{ ";for(auto itr=v.begin();itr!=v.end();++itr) {if (itr!=v.begin()) {s<< ", ";}s<<(*itr);}s<<" }";return s;}

template<typename T1, typename T2> ostream& operator<<(ostream& s,const map<T1,T2>& m){s<<"{"<<endl;for(auto itr=m.begin();itr!=m.end();++itr){s<<" "<<(*itr).first<<" : "<<(*itr).second<<endl;}s<<"}"<<endl;return s;}



const ll mod=1e9+7;

const ll inf=1e17;

const int INF=1e9;

const double EPS=1e-10;

const double PI=acos(-1);



template <class T>

map<T,T> prime_factor(T x){

	map<T,T> ans;

	for(T i=2;i*i<=x;i++){

		while(x%i==0){

			x/=i;

			ans[i]++;

		}

	}

	if (x!=1){

		ans[x]=1;

	}

	return ans;

}



//エラトステネスの篩,n以下の素数の列挙の時に

//primeはi番目の素数、適当に返却を変えて

template <class T>

vector<ll> sieve(T n){

    vector<bool> is_prime(n+1,true);

    vector<ll> prime;

    is_prime[0]=false;

    is_prime[1]=false;

    T ret=0;

    FOR(i,2,n+1){

        if(is_prime[i]){

            prime.push_back(i);

            for(T j=2*i;j<=n;j+=i){

                is_prime[j]=false;

            }

        }

    }

    return prime;

}



int main(){

    cin.tie(0);ios::sync_with_stdio(false);

    ll n;

    cin>>n;

    ll ans=0;

    vector<ll> cnt(n+1,1);

    vector<ll> ret=sieve(n);

    for(auto p:ret){

        for(int i=p;i<=n;i+=p){

            ll now=0;

            ll num=i;

            while(num%p==0){

                now++;

                num/=p;

            }

            cnt[i]*=(now+1);

        }

    }

    FOR(k,1,n+1){

        ans+=k*cnt[k];

    }

    cout<<ans<<endl;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long double ld;

typedef pair<ll,ll> P;

typedef pair<int,int> Pi;

#define rep(i,n) for(ll i=0;i<n;i++)

#define FOR(i,a,b) for(ll i=a;i<b;i++)

#define fi first

#define se second

#define endl "\n"



template<typename T> inline bool chmax(T &a, T b){if(a<b){a=b;return true;}return false;}

template<typename T> inline bool chmin(T &a, T b){if(a>b){a=b;return true;}return false;}

template<typename T> ostream& operator<<(ostream& s,const complex<T>& d) {return s<<"("<<d.real()<<", "<<d.imag()<< ")";}

template<typename T1, typename T2> ostream& operator<<(ostream& s,const pair<T1,T2>& d) {return s<<"("<<d.first<<", "<<d.second<<")";}

template<typename T> ostream& operator<<(ostream& s, const vector<T>& d){int len=d.size();rep(i,len){s<<d[i];if(i<len-1) s<<" ";}return s;}

template<typename T> ostream& operator<<(ostream& s,const vector<vector<T>>& d){int len=d.size();rep(i,len){s<<d[i]<<endl;}return s;}

template<typename T> ostream& operator<<(ostream& s,const set<T>& v){s<<"{ ";for(auto itr=v.begin();itr!=v.end();++itr) {if (itr!=v.begin()) {s<< ", ";}s<<(*itr);}s<<" }";return s;}

template<typename T> ostream& operator<<(ostream& s,const multiset<T>& v){s<<"{ ";for(auto itr=v.begin();itr!=v.end();++itr) {if (itr!=v.begin()) {s<< ", ";}s<<(*itr);}s<<" }";return s;}

template<typename T1, typename T2> ostream& operator<<(ostream& s,const map<T1,T2>& m){s<<"{"<<endl;for(auto itr=m.begin();itr!=m.end();++itr){s<<" "<<(*itr).first<<" : "<<(*itr).second<<endl;}s<<"}"<<endl;return s;}



const ll mod=1e9+7;

const ll inf=1e17;

const int INF=1e9;

const double EPS=1e-10;

const double PI=acos(-1);



int main(){

    cin.tie(0);ios::sync_with_stdio(false);

    ll n;

    cin>>n;

    ll ans=0;

    vector<ll> cnt(n+1,0);

    FOR(i,1,n+1){

        for(int j=i;j<=n;j+=i){

            cnt[j]++;

        }

    }

    FOR(k,1,n+1){

        ans+=k*cnt[k];

    }

    cout<<ans<<endl;

}

// Fastest known solution for this problem: