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: Wrong Answer)

/*{{{*/
#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define vi vector<int>
#define vl vector<ll>
#define pii pair<int,int>
#define pll pair<ll,ll>
// pll pp; -> (pp.first,pp.second)
// vector<vl> pp(n,vl(2)); sort(pp.begin(),pp.end(),[](vl a,vl b){return a[0]<b[0];});

#define REP(i,n) for(ll (i)=0;(i)<(n);(i)++)
#define RREP(i,n) for(ll (i)=(n)-1;(i)>=0;(i)--)
#define DEBUG(x) cout << #x << ": " << x << endl
#define CHMAX(a,b) a=max((a),(b))
#define CHMIN(a,b) a=min((a),(b))
/*}}}*/
int main() {
  ll n;
  cin >> n;
  ll sq = sqrt(n);
  sq++;
  ll ans = 0ll;
  ll mn=0;
  REP(i,sq){
    ll m = i+1ll;
    if(n/m==n%m) {
      cout << m << endl;
      ans += m;
      mn = n/m;
    }
  }
  if(mn==0) mn = n/sq+1;
  REP(i,mn-1){
    ll k = i+1;
    ll m = (n-k)/k;
    if(n/m==n%m) {
      ans += m;
    }
  }
  cout << ans << endl;
  return 0;
}