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<vector>
#include<string>
#include<cmath>
#include<algorithm>
#include<functional>
#include<numeric>
#include<queue>
#include<stack>
#include<map>
#include<unordered_map>
#include<set>
#include<bitset>
#include<random>

#pragma region
using namespace std;
#define FOR(i,r,n) for(ll i = (ll)(r); i < (ll)(n); i++)
#define rep(i,n) FOR(i,0LL,n)
#define RFOR(i,r,n) for(ll i=(ll)(n-1);i>=r;i--)
#define rrep(i,n) RFOR(i,0LL,n)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define COUNT(a,y,x) upper_bound(all(a), y) - lower_bound(all(a), x)
#define UNIQUE(a) sort(all(a)); a.erase(unique(all(a)), a.end())
#define pb push_back
typedef long long int ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef map<ll, ll> MAP;
const ll inf = 1LL << 61;
const ll mod = 1000000007LL;
//const ll mod = 998244353LL;

ll n = 0, m = 0, ans = 0, tmp = 0, ma = -inf, mi = inf;
string s;
bool ok;
ll dx[9] = { 0,1,0,-1,0,1,1,-1,-1 }, dy[9] = { 0,0,1,0,-1,1,-1,1,-1 };
#define endl '\n'
#pragma endregion
#define MAX 222222






int main(void) {
  ios::sync_with_stdio(false); cin.tie(0);


  cin >> n;

  if (n < MAX*10) {
    ll x = 0;
    FOR(i, 1, n + 1) {
      if (n / i == n % i) x += i;
    }
    cout << x << endl;
    return 0;
  }


  FOR(i, 1, sqrt(n) + 1) {
    if (i*i == n) continue;
    if (n%i == 0) {
      ans += (n / i) - 1;
    }
  }
  cout << ans << endl;



  return 0;
}