#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, t, x, i, flag = 0, count, z = 0;
  cin >> n;
  pair<long long int, long long int> p[100000];
  count = pow(n, 1.0 / 3);
  if (pow(count, 3) != n) count++;
  for (i = 1; i <= 2000000; i++) {
    t = i * (i - 1) * (i + 1) / 6;
    t = n + t;
    x = i * (i + 1) / 2;
    if (t % x == 0 && t / x >= i) {
      if (i == t / x) flag++;
      p[++z] = {i, t / x};
    }
  }
  if (n > 2)
    cout << 2 * z - flag << endl;
  else
    cout << n << endl;
  for (int j = 1; j <= z; j++) {
    cout << p[j].first << " " << p[j].second << endl;
  }
  if (n > 1) {
    for (int j = z; j > 0; j--) {
      if (p[j].first != p[j].second)
        cout << p[j].second << " " << p[j].first << endl;
    }
  }
}

### 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 int n, t, x, i, flag = 0, count, z = 0;
  cin >> n;
  pair<long long int, long long int> p[100000];
  count = pow(n, 1.0 / 3);
  if (pow(count, 3) != n) count++;
  for (i = 1; i <= 2000000; i++) {
    t = i * (i - 1) * (i + 1) / 6;
    t = n + t;
    x = i * (i + 1) / 2;
    if (t % x == 0 && t / x >= i) {
      if (i == t / x) flag++;
      p[++z] = {i, t / x};
    }
  }
  if (n > 2)
    cout << 2 * z - flag << endl;
  else
    cout << n << endl;
  for (int j = 1; j <= z; j++) {
    cout << p[j].first << " " << p[j].second << endl;
  }
  if (n > 1) {
    for (int j = z; j > 0; j--) {
      if (p[j].first != p[j].second)
        cout << p[j].second << " " << p[j].first << endl;
    }
  }
}

```