#include <bits/stdc++.h>
using namespace std;
int main() {
  unsigned long long X, p, q, num, den;
  vector<pair<unsigned long long, unsigned long long>> a;
  cin >> X;
  for (p = 1; p <= 6000000; p++) {
    num = 6ULL * X + p * (p * p - 1ULL);
    den = 3ULL * p * (p + 1ULL);
    q = num / den;
    if (q * den == num) {
      if (p <= q) {
        a.push_back(make_pair(p, q));
        if (p != q) a.push_back(make_pair(q, p));
      }
    }
  }
  cout << a.size() << endl;
  ;
  sort(a.begin(), a.end());
  for (long long i = 0; i < a.size(); i++) {
    cout << a[i].first << " " << a[i].second << endl;
  }
  return 0;
}

### 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;
int main() {
  unsigned long long X, p, q, num, den;
  vector<pair<unsigned long long, unsigned long long>> a;
  cin >> X;
  for (p = 1; p <= 6000000; p++) {
    num = 6ULL * X + p * (p * p - 1ULL);
    den = 3ULL * p * (p + 1ULL);
    q = num / den;
    if (q * den == num) {
      if (p <= q) {
        a.push_back(make_pair(p, q));
        if (p != q) a.push_back(make_pair(q, p));
      }
    }
  }
  cout << a.size() << endl;
  ;
  sort(a.begin(), a.end());
  for (long long i = 0; i < a.size(); i++) {
    cout << a[i].first << " " << a[i].second << endl;
  }
  return 0;
}

```