#include <bits/stdc++.h>
using namespace std;
long long int func1(long long int i) {
  return i * (i + 1) * (2 * i + 1) / 6 - (i + i + 2) * i * (i + 1) / 2 +
         (i + i + 1 + i * i) * i;
}
long long int func2(long long int i) { return i + i * i - i * (i + 1) / 2; }
int main() {
  long long int inp;
  cin >> inp;
  set<pair<long long int, long long int> > s;
  for (long long int i = 1; func1(i) <= inp; i++)
    if ((inp - func1(i)) % func2(i) == 0) {
      s.insert(make_pair(i, i + (inp - func1(i)) / func2(i)));
      s.insert(make_pair(i + (inp - func1(i)) / func2(i), i));
    }
  cout << s.size() << '\n';
  for (auto i = s.begin(); i != s.end(); i++)
    cout << (*i).first << " " << (*i).second << '\n';
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int func1(long long int i) {
  return i * (i + 1) * (2 * i + 1) / 6 - (i + i + 2) * i * (i + 1) / 2 +
         (i + i + 1 + i * i) * i;
}
long long int func2(long long int i) { return i + i * i - i * (i + 1) / 2; }
int main() {
  long long int inp;
  cin >> inp;
  set<pair<long long int, long long int> > s;
  for (long long int i = 1; func1(i) <= inp; i++)
    if ((inp - func1(i)) % func2(i) == 0) {
      s.insert(make_pair(i, i + (inp - func1(i)) / func2(i)));
      s.insert(make_pair(i + (inp - func1(i)) / func2(i), i));
    }
  cout << s.size() << '\n';
  for (auto i = s.begin(); i != s.end(); i++)
    cout << (*i).first << " " << (*i).second << '\n';
  return 0;
}

```