#include <bits/stdc++.h>
using namespace std;
int main() {
  set<pair<long long, long long> > S;
  long long x;
  cin >> x;
  S.insert(make_pair(1, x));
  S.insert(make_pair(x, 1));
  for (long long n = 2; n * n * n <= 8 * x; ++n) {
    if ((6 * x + n * n * n - n) % (3 * n * (n + 1))) continue;
    long long m = (6 * x + n * n * n - n) / (3 * n * (n + 1));
    if (n > m) break;
    if (-n * n * n + 3 * n * n * m + 3 * n * m + n == 6 * x) {
      S.insert(make_pair(n, m));
      S.insert(make_pair(m, n));
    }
  }
  cout << S.size() << endl;
  for (set<pair<long long, long long> >::iterator it = S.begin(); it != S.end();
       ++it)
    cout << it->first << " " << it->second << endl;
  return 0;
}

### Prompt

Your task is to create 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;
int main() {
  set<pair<long long, long long> > S;
  long long x;
  cin >> x;
  S.insert(make_pair(1, x));
  S.insert(make_pair(x, 1));
  for (long long n = 2; n * n * n <= 8 * x; ++n) {
    if ((6 * x + n * n * n - n) % (3 * n * (n + 1))) continue;
    long long m = (6 * x + n * n * n - n) / (3 * n * (n + 1));
    if (n > m) break;
    if (-n * n * n + 3 * n * n * m + 3 * n * m + n == 6 * x) {
      S.insert(make_pair(n, m));
      S.insert(make_pair(m, n));
    }
  }
  cout << S.size() << endl;
  for (set<pair<long long, long long> >::iterator it = S.begin(); it != S.end();
       ++it)
    cout << it->first << " " << it->second << endl;
  return 0;
}

```