#include <bits/stdc++.h>
using namespace std;
long long n;
set<long long> r;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  r.insert(1);
  for (long long m = 1; m * m <= n; m++)
    if (n % m == 0) {
      r.insert(n / m + m * ((n / m - 1) * (n / m) / 2));
      long long mm = n / m;
      r.insert(n / mm + mm * ((n / mm - 1) * (n / mm) / 2));
    }
  for (long long x : r) cout << x << " ";
  cout << "\n";
}

### Prompt

Create a solution in Cpp for the following problem:

There are n people sitting in a circle, numbered from 1 to n in the order in which they are seated. That is, for all i from 1 to n-1, the people with id i and i+1 are adjacent. People with id n and 1 are adjacent as well.

The person with id 1 initially has a ball. He picks a positive integer k at most n, and passes the ball to his k-th neighbour in the direction of increasing ids, that person passes the ball to his k-th neighbour in the same direction, and so on until the person with the id 1 gets the ball back. When he gets it back, people do not pass the ball any more.

For instance, if n = 6 and k = 4, the ball is passed in order [1, 5, 3, 1]. 

Consider the set of all people that touched the ball. The fun value of the game is the sum of the ids of people that touched it. In the above example, the fun value would be 1 + 5 + 3 = 9.

Find and report the set of possible fun values for all choices of positive integer k. It can be shown that under the constraints of the problem, the ball always gets back to the 1-st player after finitely many steps, and there are no more than 10^5 possible fun values for given n.

Input

The only line consists of a single integer n (2 ≤ n ≤ 10^9) — the number of people playing with the ball.

Output

Suppose the set of all fun values is f_1, f_2, ..., f_m.

Output a single line containing m space separated integers f_1 through f_m in increasing order.

Examples

Input


6


Output


1 5 9 21


Input


16


Output


1 10 28 64 136

Note

In the first sample, we've already shown that picking k = 4 yields fun value 9, as does k = 2. Picking k = 6 results in fun value of 1. For k = 3 we get fun value 5 and with k = 1 or k = 5 we get 21. 

<image>

In the second sample, the values 1, 10, 28, 64 and 136 are achieved for instance for k = 16, 8, 4, 10 and 11, respectively.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n;
set<long long> r;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  r.insert(1);
  for (long long m = 1; m * m <= n; m++)
    if (n % m == 0) {
      r.insert(n / m + m * ((n / m - 1) * (n / m) / 2));
      long long mm = n / m;
      r.insert(n / mm + mm * ((n / mm - 1) * (n / mm) / 2));
    }
  for (long long x : r) cout << x << " ";
  cout << "\n";
}

```