#include <bits/stdc++.h>
using namespace std;
long long X;
long long Ans[8000010][2] = {{0}};
int Ansp = 0;
int main() {
  cin >> X;
  X *= 6;
  for (long long i = 1;; i++) {
    if (i * i * i > X) break;
    long long div = i * (i + 1);
    if (X % div == 0) {
      long long MM = X / div;
      long long M3 = MM + i - 1;
      if (M3 % 3 != 0) continue;
      if (M3 / 3 < i) break;
      Ansp++;
      Ans[Ansp][0] = i, Ans[Ansp][1] = M3 / 3;
    }
  }
  for (int i = Ansp; i >= 1; i--) {
    if (Ans[i][0] != Ans[i][1]) {
      Ansp++;
      Ans[Ansp][0] = Ans[i][1];
      Ans[Ansp][1] = Ans[i][0];
    }
  }
  cout << Ansp << endl;
  for (int i = 1; i <= Ansp; i++) printf("%I64d %I64d\n", Ans[i][0], Ans[i][1]);
  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;
long long X;
long long Ans[8000010][2] = {{0}};
int Ansp = 0;
int main() {
  cin >> X;
  X *= 6;
  for (long long i = 1;; i++) {
    if (i * i * i > X) break;
    long long div = i * (i + 1);
    if (X % div == 0) {
      long long MM = X / div;
      long long M3 = MM + i - 1;
      if (M3 % 3 != 0) continue;
      if (M3 / 3 < i) break;
      Ansp++;
      Ans[Ansp][0] = i, Ans[Ansp][1] = M3 / 3;
    }
  }
  for (int i = Ansp; i >= 1; i--) {
    if (Ans[i][0] != Ans[i][1]) {
      Ansp++;
      Ans[Ansp][0] = Ans[i][1];
      Ans[Ansp][1] = Ans[i][0];
    }
  }
  cout << Ansp << endl;
  for (int i = 1; i <= Ansp; i++) printf("%I64d %I64d\n", Ans[i][0], Ans[i][1]);
  return 0;
}

```