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;
int main() {
  unsigned long long x;
  cin >> x;
  vector<pair<unsigned long long, unsigned long long> > res;
  for (unsigned long long n = 1; n <= 1000000 + 500000; n++) {
    if ((6 * x) % (n * (n + 1)) == 0 &&
        (6 * x / (n * (n + 1)) + n - 1) % 3 == 0) {
      unsigned long long m = (6 * x / (n * (n + 1)) + n - 1) / 3;
      if (m >= n) {
        res.push_back({n, m});
        res.push_back({m, n});
      }
    }
  }
  sort(res.begin(), res.end());
  int cur = (int)res.size();
  for (int i = 1; i < (int)res.size(); i++)
    if (res[i] == res[i - 1]) cur--;
  cout << cur << endl;
  for (int i = 0; i < (int)res.size(); i++) {
    if (!i)
      cout << res[i].first << " " << res[i].second << endl;
    else if (i && res[i] != res[i - 1])
      cout << res[i].first << " " << res[i].second << endl;
  }
  return 0;
}