#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x, m, n;
  cin >> x;
  x *= 6ll;
  long long equ = 0;
  vector<pair<long long, long long> > v;
  for (m = 1; m * (m + 1) <= x / (2 * m + 1); m++) {
    if (x % (m * (m + 1)) == 0 && (x / m / (m + 1) - 1 + m) % 3 == 0) {
      n = (x / m / (m + 1) + m - 1) / 3;
      if (n == m)
        equ = n;
      else if (n > m) {
        v.push_back({m, n});
      }
    }
  }
  long long cnt = 2ll * v.size() + (equ != 0);
  cout << cnt << endl;
  for (int i = 0; i < v.size(); i++)
    cout << v[i].first << ' ' << v[i].second << endl;
  if (equ != 0) cout << equ << ' ' << equ << endl;
  for (int i = v.size() - 1; i >= 0; i--)
    cout << v[i].second << ' ' << v[i].first << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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, m, n;
  cin >> x;
  x *= 6ll;
  long long equ = 0;
  vector<pair<long long, long long> > v;
  for (m = 1; m * (m + 1) <= x / (2 * m + 1); m++) {
    if (x % (m * (m + 1)) == 0 && (x / m / (m + 1) - 1 + m) % 3 == 0) {
      n = (x / m / (m + 1) + m - 1) / 3;
      if (n == m)
        equ = n;
      else if (n > m) {
        v.push_back({m, n});
      }
    }
  }
  long long cnt = 2ll * v.size() + (equ != 0);
  cout << cnt << endl;
  for (int i = 0; i < v.size(); i++)
    cout << v[i].first << ' ' << v[i].second << endl;
  if (equ != 0) cout << equ << ' ' << equ << endl;
  for (int i = v.size() - 1; i >= 0; i--)
    cout << v[i].second << ' ' << v[i].first << endl;
  return 0;
}

```