#include <bits/stdc++.h>
using namespace std;
vector<unsigned long long> ans;
int main() {
  unsigned long long X;
  cin >> X;
  for (unsigned long long i = 1; i * (i + 1) * (2 * i + 1) <= 6 * X; i++) {
    if (X * 6 % (i * i + i) != 0) continue;
    if ((X * 6 / (i * i + i) + i - 1) % 3 == 0) {
      ans.push_back(i);
    }
  }
  int last = ans.size() - 1;
  while (last >= 0) {
    unsigned long long i = ans[last--];
    unsigned long long j = (X * 6 / i / (i + 1) + i - 1) / 3;
    if (i != j) ans.push_back(j);
  }
  printf("%d\n", ans.size());
  for (int i = 0, sz = ans.size() - 1; i <= sz; i++) {
    cout << ans[i] << " " << ans[sz - i] << endl;
  }
}

### Prompt

Generate 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;
vector<unsigned long long> ans;
int main() {
  unsigned long long X;
  cin >> X;
  for (unsigned long long i = 1; i * (i + 1) * (2 * i + 1) <= 6 * X; i++) {
    if (X * 6 % (i * i + i) != 0) continue;
    if ((X * 6 / (i * i + i) + i - 1) % 3 == 0) {
      ans.push_back(i);
    }
  }
  int last = ans.size() - 1;
  while (last >= 0) {
    unsigned long long i = ans[last--];
    unsigned long long j = (X * 6 / i / (i + 1) + i - 1) / 3;
    if (i != j) ans.push_back(j);
  }
  printf("%d\n", ans.size());
  for (int i = 0, sz = ans.size() - 1; i <= sz; i++) {
    cout << ans[i] << " " << ans[sz - i] << endl;
  }
}

```