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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <tuple>
#include <stack>
#include <bitset>
#include <utility>
// #include <numeric>
using namespace std;



typedef unsigned long ul;
typedef unsigned long long ull;
typedef long long ll;
typedef vector<ll> vint;
typedef vector<vector<ll>> vvint;
typedef pair<ll,ll> pint;
typedef vector<pint> vpint;

#define rep(i,n) for(ll i=0;i<n;i++)
#define repf(i,f,n) for(ll i=f;i<n;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define INT_MAX 2147483647
#define LLONG_MAX 9223372036854775807
#define vmax(vec) *max_element(vec.begin(), vec.end())
#define vmin(vec) *min_element(vec.begin(), vec.end())
#define vsort(vec) sort(vec.begin(), vec.end())
#define vsortgr(vec) sort(vec.begin(), vec.end(), greater<ll>())


typedef struct{
  ll pos;
  ll depth;
} data;
//
// struct Node{
//
// }



int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0,};

// ll X,Y,Z,K;
// ll pmax = 30000000000;
// vint A,B,C;




// int binary_search(int key){
//  //isOKを満たす要素のうち最小のインデックスを返す。（要ソート）
//   int left = -1;
//   int right = N;
//
//   while(right - left > 1){
//     int mid = left + (right - left)/2;
//
//     if(isOK(mid, key)) right=mid;
//     else left=mid;
//   }
//
//   return right;
// }

// ll gcd(ll m, ll n){
//   if(m<n) swap(m,n);
//   if(n==0) return m;
//   return gcd(n, m%n);
// }

// nの約数のうち2n未満のもの内、n/i-1の和を返す。
ll divisor(ll n){
  ll val = 0;

  for(ll i=1; i*i < n ; i++){
    // if(i>2*n) break;
    if(n%i==0 && (n/i-1)>1){
      // cout << i << " " << (n/i-1) << endl;
      val += (n/i-1);
    }
  }

  return val;

}



int main() {

  ll N;
  cin >> N;
  ll ans=0;

  // ll n = 1;
  // while(N>2*n){
  //   if((N-n)%n==0){
  //     ans+=(N-n)/n;
  //   }
  //
  //   n++;
  // }

  ans = divisor(N);

  cout << ans << endl;
}

















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