#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long x;
    cin >> x;
    set<pair<long long int, long long int> > ans;
    for (long long int i = 1; i < 2000005; i++) {
      long long n = i;
      long long t1 = 6 * x + (n * n * n) - n;
      long long t2 = 3 * (n * n + n);
      if (t1 % t2 != 0 || (t1 / t2) <= 0)
        continue;
      else if (t1 / t2 >= i) {
        ans.insert({t1 / t2, i});
        ans.insert({i, t1 / t2});
      }
    }
    cout << (long long int)ans.size() << '\n';
    for (auto it = ans.begin(); it != ans.end(); it++)
      cout << it->first << " " << it->second << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long x;
    cin >> x;
    set<pair<long long int, long long int> > ans;
    for (long long int i = 1; i < 2000005; i++) {
      long long n = i;
      long long t1 = 6 * x + (n * n * n) - n;
      long long t2 = 3 * (n * n + n);
      if (t1 % t2 != 0 || (t1 / t2) <= 0)
        continue;
      else if (t1 / t2 >= i) {
        ans.insert({t1 / t2, i});
        ans.insert({i, t1 / t2});
      }
    }
    cout << (long long int)ans.size() << '\n';
    for (auto it = ans.begin(); it != ans.end(); it++)
      cout << it->first << " " << it->second << '\n';
  }
  return 0;
}

```