#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x;
  cin >> x;
  long long counter = 0;
  vector<pair<long long, long long>> vec;
  for (long long n = 1; n <= 10000000; n++) {
    long long deno = n * (n + 1);
    deno >>= 1;
    if (deno % 3 == 0) {
      deno /= 3;
      if (x % deno == 0) {
        long long val = (x / deno);
        val += n;
        val -= 1;
        if (val > 0 && val % 3 == 0) {
          long long m = val / 3;
          if (m > n) {
            counter += 2;
            vec.push_back({n, m});
          } else if (m == n) {
            counter++;
            vec.push_back({n, n});
            break;
          }
        }
      }
    } else if (x % deno == 0) {
      long long val = 3 * (x / deno);
      val += n;
      val -= 1;
      if (val > 0 && val % 3 == 0) {
        long long m = val / 3;
        if (m > n) {
          counter += 2;
          vec.push_back({n, m});
        } else if (m == n) {
          counter++;
          vec.push_back({n, n});
          break;
        }
      }
    }
  }
  cout << counter << endl;
  if (counter % 2 == 0) {
    for (auto x : vec) {
      cout << x.first << " " << x.second << endl;
    }
    reverse(vec.begin(), vec.end());
    for (auto x : vec) {
      cout << x.second << " " << x.first << endl;
    }
  } else {
    for (auto x : vec) {
      cout << x.first << " " << x.second << endl;
    }
    vec.pop_back();
    reverse(vec.begin(), vec.end());
    for (auto x : vec) {
      cout << x.second << " " << x.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 x;
  cin >> x;
  long long counter = 0;
  vector<pair<long long, long long>> vec;
  for (long long n = 1; n <= 10000000; n++) {
    long long deno = n * (n + 1);
    deno >>= 1;
    if (deno % 3 == 0) {
      deno /= 3;
      if (x % deno == 0) {
        long long val = (x / deno);
        val += n;
        val -= 1;
        if (val > 0 && val % 3 == 0) {
          long long m = val / 3;
          if (m > n) {
            counter += 2;
            vec.push_back({n, m});
          } else if (m == n) {
            counter++;
            vec.push_back({n, n});
            break;
          }
        }
      }
    } else if (x % deno == 0) {
      long long val = 3 * (x / deno);
      val += n;
      val -= 1;
      if (val > 0 && val % 3 == 0) {
        long long m = val / 3;
        if (m > n) {
          counter += 2;
          vec.push_back({n, m});
        } else if (m == n) {
          counter++;
          vec.push_back({n, n});
          break;
        }
      }
    }
  }
  cout << counter << endl;
  if (counter % 2 == 0) {
    for (auto x : vec) {
      cout << x.first << " " << x.second << endl;
    }
    reverse(vec.begin(), vec.end());
    for (auto x : vec) {
      cout << x.second << " " << x.first << endl;
    }
  } else {
    for (auto x : vec) {
      cout << x.first << " " << x.second << endl;
    }
    vec.pop_back();
    reverse(vec.begin(), vec.end());
    for (auto x : vec) {
      cout << x.second << " " << x.first << endl;
    }
  }
}

```