#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long k;
  cin >> k;
  vector<pair<long long, long long>> ans;
  for (long long y = 1;; y++) {
    if (k < (2 * y * y * y + 3 * y * y + y) / 6) break;
    long long s = y, a = y + 1, e, f, x;
    e = (s * a - s * (s + 1) / 2);
    f = k + a * s * (s + 1) / 2 - (2 * s * s * s + 3 * s * s + s) / 6;
    if (f < 0 || f % e) continue;
    x = f / e;
    ans.push_back(make_pair(x - 1, y));
    if (x - 1 != y) {
      ans.push_back(make_pair(y, x - 1));
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < (int)(ans.size()); i++)
    cout << ans[i].first << ' ' << ans[i].second << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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::sync_with_stdio(0);
  cin.tie(0);
  long long k;
  cin >> k;
  vector<pair<long long, long long>> ans;
  for (long long y = 1;; y++) {
    if (k < (2 * y * y * y + 3 * y * y + y) / 6) break;
    long long s = y, a = y + 1, e, f, x;
    e = (s * a - s * (s + 1) / 2);
    f = k + a * s * (s + 1) / 2 - (2 * s * s * s + 3 * s * s + s) / 6;
    if (f < 0 || f % e) continue;
    x = f / e;
    ans.push_back(make_pair(x - 1, y));
    if (x - 1 != y) {
      ans.push_back(make_pair(y, x - 1));
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < (int)(ans.size()); i++)
    cout << ans[i].first << ' ' << ans[i].second << endl;
  return 0;
}

```