//素因数分解による解法も試してみる
//こちらもO(sqrt(N))くらい。
#include <iostream>
#include <vector>
#define int long long
using namespace std;

int n;

signed main() {
	typedef pair<int, int> P;
	vector<P> pr;
	
	cin >> n;
	
	for (int i = 2; i * i <= n; i++) {
		int cnt = 0;
		while (n % i == 0) {
			n /= i;
			cnt++;
		}
		if (cnt > 0) pr.push_back(P(i, cnt));
	}
	if (n > 1) pr.push_back(P(n, 1));
	
	int ans = 1;
	for (int i = 0; i < pr.size(); i++) {
		int a = pr[i].first;
		int ssum = 0;
		int mul = 1;
		for (int j = 0; j <= pr[i].second; j++) {
			ssum += mul;
			mul *= a;
		}
		ans *= ssum;
	}
	
	cout << ans << endl;
	return 0;
}