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.

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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;
const unsigned long long MAXN = (unsigned long long)1e18;
vector<pair<unsigned long long, unsigned long long> > ans;
int main() {
  unsigned long long x;
  cin >> x;
  unsigned long long MAXN = (unsigned long long)1e7;
  for (unsigned long long n = 1ll; n <= MAXN; n++) {
    if ((n * n * n + 6 * x - n) % (3 * (n * n + n)) == 0 &&
        (n * n * n + 6 * x - n) / (3 * (n * n + n)) > 0 &&
        n <= (n * n * n + 6 * x - n) / (3 * (n * n + n))) {
      ans.push_back(make_pair(n, (n * n * n + 6 * x - n) / (3 * (n * n + n))));
      if (n != (n * n * n + 6 * x - n) / (3 * (n * n + n)))
        ans.push_back(
            make_pair((n * n * n + 6 * x - n) / (3 * (n * n + n)), n));
    }
  }
  cout << ans.size() << endl;
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].first << ' ' << ans[i].second << endl;
  }
  return 0;
}