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 <queue>
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <map>
#include <stdio.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for(int i = (a); i < (int)(b); i++)
typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vint;
typedef vector<string> vstr;
typedef pair<ll, ll> pint;
unsigned long long inf = (1LL << 62);
ll MOD = (ll)pow(10.0, 9.0) + 7;
pint dx[4] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };
char da[4] = { 'R', 'L', 'D', 'U' };
char dad[4] = { 'L', 'R', 'U', 'D' };
ll min(ll a, ll b)
{

	return (a < b) ? a : b;
}
ll max(ll a, ll b) { return (a > b) ? a : b; }

ll N;





string S[2 * 10000];
int main() {
	cin >> N;

   unsigned long long sum = 0;
	for (ll i = 1; i * i < N; i++) {
		if (N % i == 0 && ((N / i) - 1) > i) {
			
			sum += (N / i) - 1;
		}
	}

	if (N == 2) {
		cout << 0 << endl;
		return 0;
	}

	cout << sum << endl;
	return 0;

}