#include <bits/stdc++.h>
using namespace std;
long long k, x;
set<pair<long long, long long> > s;
int main() {
  cin >> x;
  for (long long m = 1; m * m * m <= 6 * x; m++) {
    if ((6 * x) % m == 0) {
      long long a = (6 * x) / m + m * m - 1;
      if (a % (3 * m + 3) == 0) {
        long long n = a / (3 * m + 3);
        if (m <= n) {
          s.insert(make_pair(m, n));
          if (n != m) s.insert(make_pair(n, m));
        }
      }
    }
  }
  k = s.size();
  cout << k << "\n";
  for (long long i = 0; i < k; i++) {
    cout << (*(s.begin())).first << " " << (*(s.begin())).second << "\n";
    s.erase(s.begin());
  }
}

### Prompt

Please create a solution in cpp 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;
long long k, x;
set<pair<long long, long long> > s;
int main() {
  cin >> x;
  for (long long m = 1; m * m * m <= 6 * x; m++) {
    if ((6 * x) % m == 0) {
      long long a = (6 * x) / m + m * m - 1;
      if (a % (3 * m + 3) == 0) {
        long long n = a / (3 * m + 3);
        if (m <= n) {
          s.insert(make_pair(m, n));
          if (n != m) s.insert(make_pair(n, m));
        }
      }
    }
  }
  k = s.size();
  cout << k << "\n";
  for (long long i = 0; i < k; i++) {
    cout << (*(s.begin())).first << " " << (*(s.begin())).second << "\n";
    s.erase(s.begin());
  }
}

```