Problem:
Score : 500 points

Problem Statement

Snuke received a positive integer N from Takahashi.
A positive integer m is called a favorite number when the following condition is satisfied:

The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds.

Find all favorite numbers and print the sum of those.

Constraints

All values in input are integers.

1 \leq N \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

8

Sample Output 1

10

There are two favorite numbers: 3 and 7. Print the sum of these, 10.

Sample Input 2

1000000000000

Sample Output 2

2499686339916

Watch out for overflow.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <sstream>
#include <stack>
#include <iomanip>
#include <numeric>
#include <string>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cstdlib>
#include <cassert>
#include <map>
#include <set>
#include <vector>
#include <utility>
#include <queue>
#include <cmath>

#define BEGIN(x) x.begin()
#define END(x) x.end()
#define ALL(x) BEGIN(x), END(x)
#define PAIR make_pair
#define VEC(type) vector<type >
#define endl '\n'
#define FIR first
#define SEC second
#define PB push_back
#define SZ(x) ((int)x.size())

#define FOR(i, A, B) for(auto i = (A);i != (B); i++)
#define FORD(i, A, B) for(auto i = (A);i != (B); i--)

#define READRANGE(begin, end) FOR(it, begin, end) cin >> *it
#define READVEC(V) READRANGE(BEGIN(V), END(V))

using namespace std;

typedef long long lint;
typedef pair<int, int> Pii;
typedef pair<int, lint> Pil;
typedef pair<lint, lint> Pll;
typedef pair<lint, int> Pli;

template<class U, class V>
istream& operator >>(istream& in, pair<U, V>& pr) {
    return in >> pr.first >> pr.second;
}

template<class U, class V>
ostream& operator <<(ostream& out, pair<U, V>& p) {
    return out << p.first << " " << p.second;
}

template<class T>
ostream& operator <<(ostream& out, vector<T> vec) {
    for (T& e : vec)
        out << e << " ";
    return out << endl;
}

// ---- BEGIN LIBRARY CODE ----
// ---- END LIBRARY CODE ----
void io_init()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
};

void solveSingleCase(int caseNo)
{
    lint N;
    cin >> N;

    set<lint> favs;
    for (lint p = 1;p <= N / p; p++)
        if (N % p == 0) {
            lint m = N / p - 1;
            if (m > 0 && N / m == N % m)
                favs.insert(m);
            m = p - 1;
            if (m > 0 && N / m == N % m)
                favs.insert(m);
        }

    lint total = accumulate(ALL(favs), 0LL);
    cout << total << endl;
};

int main(int argc, char **argv)
{
    io_init();

    // Actual program code starts here.
    int T = 1;
    // cin >> T;
    FOR(tcase, 1, T + 1) {
        solveSingleCase(tcase);
    }
    return 0;
};