#include <bits/stdc++.h>
using namespace std;
set<pair<long long, long long> > sol;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  long long N;
  cin >> N;
  for (long long i = 1;; i++) {
    long long a = (6ll * N + (i * i * i) - i);
    long long b = 3ll * (i * i) + 3 * i;
    long long x = a / b;
    if (x < i) break;
    if (a % b != 0) continue;
    sol.insert(pair<long long, long long>(i, x));
    sol.insert(pair<long long, long long>(x, i));
  }
  cout << sol.size() << '\n';
  if (sol.size() != 0) {
    auto it = sol.begin();
    cout << it->first << " " << it->second;
    it++;
    for (; it != sol.end(); it++)
      cout << "\n" << it->first << " " << it->second;
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

Spongebob is already tired trying to reason his weird actions and calculations, so he simply asked you to find all pairs of n and m, such that there are exactly x distinct squares in the table consisting of n rows and m columns. For example, in a 3 × 5 table there are 15 squares with side one, 8 squares with side two and 3 squares with side three. The total number of distinct squares in a 3 × 5 table is 15 + 8 + 3 = 26.

Input

The first line of the input contains a single integer x (1 ≤ x ≤ 1018) — the number of squares inside the tables Spongebob is interested in.

Output

First print a single integer k — the number of tables with exactly x distinct squares inside.

Then print k pairs of integers describing the tables. Print the pairs in the order of increasing n, and in case of equality — in the order of increasing m.

Examples

Input

26


Output

6
1 26
2 9
3 5
5 3
9 2
26 1


Input

2


Output

2
1 2
2 1


Input

8


Output

4
1 8
2 3
3 2
8 1

Note

In a 1 × 2 table there are 2 1 × 1 squares. So, 2 distinct squares in total.

<image>

In a 2 × 3 table there are 6 1 × 1 squares and 2 2 × 2 squares. That is equal to 8 squares in total.

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
set<pair<long long, long long> > sol;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  long long N;
  cin >> N;
  for (long long i = 1;; i++) {
    long long a = (6ll * N + (i * i * i) - i);
    long long b = 3ll * (i * i) + 3 * i;
    long long x = a / b;
    if (x < i) break;
    if (a % b != 0) continue;
    sol.insert(pair<long long, long long>(i, x));
    sol.insert(pair<long long, long long>(x, i));
  }
  cout << sol.size() << '\n';
  if (sol.size() != 0) {
    auto it = sol.begin();
    cout << it->first << " " << it->second;
    it++;
    for (; it != sol.end(); it++)
      cout << "\n" << it->first << " " << it->second;
  }
  return 0;
}

```