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;
vector<pair<long long, long long> > ans;
vector<pair<long long, long long> > ans2;
int main() {
  std::ios_base::sync_with_stdio(false);
  unsigned long long x;
  cin >> x;
  int limit;
  long double rad = x * 1.0;
  rad = pow(rad, 1.0 / 3.0);
  rad = rad * 2.0;
  limit = 2000000;
  ans.push_back(make_pair(1LL, x));
  for (int i = 1; i <= limit; i++) {
    unsigned long long nr = 6LL * x;
    unsigned long long nr2 = 1LL * i * i * i;
    unsigned long long nr3 = nr + nr2 - i;
    unsigned long long nr4 = 3LL * i * i + 3LL * i;
    if (nr3 % nr4 == 0 && i <= nr3 / nr4) {
      ans.push_back(make_pair(i, nr3 / nr4));
      ans.push_back(make_pair(nr3 / nr4, i));
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); ++i) {
    if (i == 0) ans2.push_back(ans[i]);
    if (i > 0 && ans[i] != ans[i - 1]) ans2.push_back(ans[i]);
  }
  cout << ans2.size() << "\n";
  for (int i = 0; i < ans2.size(); ++i)
    cout << ans2[i].first << " " << ans2[i].second << "\n";
  return 0;
}