#include <bits/stdc++.h>
using namespace std;
const int MAX_V = 10000000;
int main() {
  long long x;
  vector<pair<long long, long long> > ans;
  cin >> x;
  long long n = 1, s_i = 0, ss_i = 0;
  while (n <= MAX_V) {
    s_i += n;
    ss_i += n * n;
    long long s1 = x + n * s_i - ss_i;
    long long s2 = (n + 1) * n - s_i;
    if (s1 % s2 == 0 && s1 >= s2) {
      long long m = s1 / s2;
      if (n > m) break;
      ans.push_back(make_pair(n, m));
      if (n != m) ans.push_back(make_pair(m, n));
    }
    n++;
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].first << " " << ans[i].second << "\n";
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int MAX_V = 10000000;
int main() {
  long long x;
  vector<pair<long long, long long> > ans;
  cin >> x;
  long long n = 1, s_i = 0, ss_i = 0;
  while (n <= MAX_V) {
    s_i += n;
    ss_i += n * n;
    long long s1 = x + n * s_i - ss_i;
    long long s2 = (n + 1) * n - s_i;
    if (s1 % s2 == 0 && s1 >= s2) {
      long long m = s1 / s2;
      if (n > m) break;
      ans.push_back(make_pair(n, m));
      if (n != m) ans.push_back(make_pair(m, n));
    }
    n++;
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].first << " " << ans[i].second << "\n";
  }
  return 0;
}

```