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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool compare(pair<long long, long long>& a, pair<long long, long long>& b) {
  if (a.first != b.first) return (a.first < b.first);
  return (a.second < b.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long X;
  cin >> X;
  vector<pair<long long, long long>> ans, ans2;
  long long n = 1;
  while (n <= (floor(2 * cbrt(X)) < floor(sqrt(X)) ? floor(2 * cbrt(X))
                                                   : floor(sqrt(X)))) {
    long long num = 6 * X + (n * n * n) - n;
    long long den = 3 * ((n * n) + n);
    if (num % den == 0 && n <= num / den) {
      ans.push_back(pair<long long, long long>(n, num / den));
      if (n != num / den)
        ans.push_back(pair<long long, long long>(num / den, n));
    }
    n++;
  }
  sort(ans.begin(), ans.end(), compare);
  cout << ans.size() << "\n";
  for (int i = int(0); i <= int(ans.size() - 1); i++)
    cout << ans[i].first << " " << ans[i].second << "\n";
}