#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long x, c = 0;
  cin >> x;
  set<pair<long long, long long> > out;
  for (long long n = 1; n <= 2000000; n++) {
    long long m = 2 * x + n * (n * n - 1) / 3;
    if (m % (n * (n + 1)) == 0) {
      m = m / n / (n + 1);
      if (n > m) continue;
      if (n == m)
        c++;
      else
        c += 2;
      out.insert({n, m});
      out.insert({m, n});
    }
  }
  cout << c << "\n";
  for (auto i = out.begin(); i != out.end(); i++) {
    cout << i->first << " " << i->second << "\n";
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long x, c = 0;
  cin >> x;
  set<pair<long long, long long> > out;
  for (long long n = 1; n <= 2000000; n++) {
    long long m = 2 * x + n * (n * n - 1) / 3;
    if (m % (n * (n + 1)) == 0) {
      m = m / n / (n + 1);
      if (n > m) continue;
      if (n == m)
        c++;
      else
        c += 2;
      out.insert({n, m});
      out.insert({m, n});
    }
  }
  cout << c << "\n";
  for (auto i = out.begin(); i != out.end(); i++) {
    cout << i->first << " " << i->second << "\n";
  }
  return 0;
}

```