#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x;
  vector<pair<long long, long long> > ans;
  cin >> x;
  long long p = cbrt(6 * x) + 2, n, m;
  for (m = 1; m <= p; m++) {
    long long a = (6 * x + m * (m * m - 1)), b = (3 * m * (m + 1));
    if (a % b == 0) {
      n = a / b;
      if (m > n)
        break;
      else
        ans.push_back({m, n});
    }
  }
  int num = 2 * ans.size(), flag = 0;
  if (ans[ans.size() - 1].first == ans[ans.size() - 1].second)
    num -= 1, flag = 1;
  printf("%d\n", num);
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << " " << ans[i].second << "\n";
  for (int i = ans.size() - 1 - flag; i >= 0; i--) {
    cout << ans[i].second << " " << ans[i].first << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write 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() {
  long long x;
  vector<pair<long long, long long> > ans;
  cin >> x;
  long long p = cbrt(6 * x) + 2, n, m;
  for (m = 1; m <= p; m++) {
    long long a = (6 * x + m * (m * m - 1)), b = (3 * m * (m + 1));
    if (a % b == 0) {
      n = a / b;
      if (m > n)
        break;
      else
        ans.push_back({m, n});
    }
  }
  int num = 2 * ans.size(), flag = 0;
  if (ans[ans.size() - 1].first == ans[ans.size() - 1].second)
    num -= 1, flag = 1;
  printf("%d\n", num);
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << " " << ans[i].second << "\n";
  for (int i = ans.size() - 1 - flag; i >= 0; i--) {
    cout << ans[i].second << " " << ans[i].first << "\n";
  }
  return 0;
}

```