// Original C++ solution (slower):
#include <iostream>

#include <iomanip>

#include<math.h>

#include<list>

#include <algorithm>

#include<set>

#include<vector>

#include<math.h>

#include<map>

#include<string>

#include <numeric>

#include <queue>

#include <sstream>

#include <bitset>

#include<stack>



using namespace std;

using ll = long long;

using vll = vector<long long>;

using sll = set<long long>;





template<typename T>

T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }

template<typename T>

T lcm(T a, T b) { return a / gcd(a, b) * b; }



template<typename T>

map<T, T> getPrimeFactor(T n) {

	map<T, T> res;

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

		while (n % i == 0) {

			res[i] ++;

			n /= i;

		}

	}

	if (n != 1) res[n] = 1;

	return res;

}



template<typename T>

bool IsPrimeNumber(T num)

{

	if (num <= 2) return true;

	else if (num % 2 == 0) return false;



	double sqrtNum = sqrt(num);

	for (int i = 3; i <= sqrtNum; i += 2)

	{

		if (num % i == 0)

		{

			return false;

		}

	}



	return true;

}



long long modinv(long long a, long long m) {

	long long b = m, u = 1, v = 0;

	while (b) {

		long long t = a / b;

		a -= t * b; swap(a, b);

		u -= t * v; swap(u, v);

	}

	u %= m;

	if (u < 0) u += m;

	return u;

}



// 繰り返し二乗法

ll pow2(ll n, ll p, ll mod) {

	if (p == 0 || n == 1) {

		return 1;

	}

	else {

		ll ret = pow2(n * n % mod, p / 2, mod);

		if (p % 2 == 1) {

			ret *= n;

		}

		return ret % mod;

	}

}



#define rep(i,s,e) for(ll i=s;i<e;i++)

#define repeq(i,s,e) for(ll i=s;i<=e;i++)





int main() {

	ll N;

	std::cin >> N;

	ll ans = 1;

	vll cnt(N+1,1);

	repeq(k, 2, N) {

		for (ll i = k; i <= N; i += k) {

			cnt[i]++;

		}

		ans += k * cnt[k];

	}



	std::cout << ans << endl;

	return 0;

}


// Optimized solution to the same problem (faster):
#include <iostream>

#include <iomanip>

#include<math.h>

#include<list>

#include <algorithm>

#include<set>

#include<vector>

#include<math.h>

#include<map>

#include<string>

#include <numeric>

#include <queue>

#include <sstream>

#include <bitset>

#include<stack>



using namespace std;

using ll = long long;

using vll = vector<long long>;

using sll = set<long long>;





template<typename T>

T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }

template<typename T>

T lcm(T a, T b) { return a / gcd(a, b) * b; }



template<typename T>

map<T, T> getPrimeFactor(T n) {

	map<T, T> res;

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

		while (n % i == 0) {

			res[i] ++;

			n /= i;

		}

	}

	if (n != 1) res[n] = 1;

	return res;

}



template<typename T>

bool IsPrimeNumber(T num)

{

	if (num <= 2) return true;

	else if (num % 2 == 0) return false;



	double sqrtNum = sqrt(num);

	for (int i = 3; i <= sqrtNum; i += 2)

	{

		if (num % i == 0)

		{

			return false;

		}

	}



	return true;

}



long long modinv(long long a, long long m) {

	long long b = m, u = 1, v = 0;

	while (b) {

		long long t = a / b;

		a -= t * b; swap(a, b);

		u -= t * v; swap(u, v);

	}

	u %= m;

	if (u < 0) u += m;

	return u;

}



// 繰り返し二乗法

ll pow2(ll n, ll p, ll mod) {

	if (p == 0 || n == 1) {

		return 1;

	}

	else {

		ll ret = pow2(n * n % mod, p / 2, mod);

		if (p % 2 == 1) {

			ret *= n;

		}

		return ret % mod;

	}

}



#define rep(i,s,e) for(ll i=s;i<e;i++)

#define repeq(i,s,e) for(ll i=s;i<=e;i++)





int main() {

	ll N;

	std::cin >> N;

	ll ans = 0;

	repeq(i, 1, N) {

		for (ll j = i; j <= N; j += i) {

			ans += j;

		}

	}

	std::cout << ans << endl;

	return 0;

}


// Fastest known solution for this problem: