#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  long long n, i, j, k, t;
  cin >> n;
  vector<pair<long long, long long> > v;
  i = 0, j = 1;
  while (i < j) {
    i++;
    j = (n - (i - 1) * i * (2 * i - 1) / 6 + i * i * (i - 1) / 2) /
        (i * i - i * (i - 1) / 2);
    if ((i * i * j - (i + j) * i * (i - 1) / 2 +
         (i - 1) * i * (2 * i - 1) / 6) != n)
      continue;
    if (i <= j) v.push_back(make_pair(i, j));
  }
  for (i = v.size() - 1; i >= 0; i--) {
    if (v[i].first != v[i].second)
      v.push_back(make_pair(v[i].second, v[i].first));
  }
  cout << v.size() << endl;
  for (i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << endl;
  return 0;
}

### Prompt

Please formulate 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() {
  ios::sync_with_stdio(false);
  long long n, i, j, k, t;
  cin >> n;
  vector<pair<long long, long long> > v;
  i = 0, j = 1;
  while (i < j) {
    i++;
    j = (n - (i - 1) * i * (2 * i - 1) / 6 + i * i * (i - 1) / 2) /
        (i * i - i * (i - 1) / 2);
    if ((i * i * j - (i + j) * i * (i - 1) / 2 +
         (i - 1) * i * (2 * i - 1) / 6) != n)
      continue;
    if (i <= j) v.push_back(make_pair(i, j));
  }
  for (i = v.size() - 1; i >= 0; i--) {
    if (v[i].first != v[i].second)
      v.push_back(make_pair(v[i].second, v[i].first));
  }
  cout << v.size() << endl;
  for (i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << endl;
  return 0;
}

```