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: Time Limit Exceeded)

#include <iostream>

using namespace std;

//#define int long long

using ll = long long;
using pii =  pair<int, int>;

const int IINF =  0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const int INF = sizeof(int) == sizeof(long long) ? LINF : IINF;
const ll MOD = 1e9+7;

#define for_(i, a, b) for(int i = (a);i < (b);++i)
#define rfor_(i, a, b) for(int i = (b)-1;i >= (a);--i)

#define rep(i, n) for_(i, 0, n)
#define rrep(i, n) rfor_(i, 0, n)

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define ft first
#define sd second

#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#define rev(a) reverse(all(a))

#define endl '\n'

const int limit = 100007;

ll N;

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

  cin >> N;

  ll sum = 0;

  for(ll k=1;k<=N/3;++k) {
    if( N % k == 0 ) {
      ll m = (N-k)/k;
      if(m<=k)break;
      sum += m;
      //cout << m << endl;
    }
  }

  cout << sum << endl;

  return 0;
}