#include <bits/stdc++.h>
using namespace std;
map<long long int, long long int> Z;
int main() {
  long long int x;
  scanf("%lld", &x);
  int k = 0;
  for (int i = 1; i <= (2 * pow(x, ((double)1) / 3)); i++) {
    long long int temp = 6 * x;
    if (temp % i == 0) {
      temp = temp / i;
      if (temp % (i + 1) == 0) {
        temp = temp / (i + 1);
        if ((temp + i - 1) % 3 == 0) {
          temp = (temp + i - 1) / 3;
          if (i <= temp) {
            k++;
            if (i != temp) k++;
            Z.insert(make_pair(i, temp));
            Z.insert(make_pair(temp, i));
          }
        }
      }
    }
  }
  printf("%d\n", k);
  for (typeof((Z).begin()) it = Z.begin(); it != (Z).end(); ++it) {
    cout << it->first << " " << it->second << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
map<long long int, long long int> Z;
int main() {
  long long int x;
  scanf("%lld", &x);
  int k = 0;
  for (int i = 1; i <= (2 * pow(x, ((double)1) / 3)); i++) {
    long long int temp = 6 * x;
    if (temp % i == 0) {
      temp = temp / i;
      if (temp % (i + 1) == 0) {
        temp = temp / (i + 1);
        if ((temp + i - 1) % 3 == 0) {
          temp = (temp + i - 1) / 3;
          if (i <= temp) {
            k++;
            if (i != temp) k++;
            Z.insert(make_pair(i, temp));
            Z.insert(make_pair(temp, i));
          }
        }
      }
    }
  }
  printf("%d\n", k);
  for (typeof((Z).begin()) it = Z.begin(); it != (Z).end(); ++it) {
    cout << it->first << " " << it->second << endl;
  }
  return 0;
}

```