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;
template <typename T, typename S>
ostream &operator<<(ostream &os, const pair<T, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename C, typename T = decay<decltype(*begin(declval<C>()))>,
          typename enable_if<!is_same<C, string>::value>::type * = nullptr>
ostream &operator<<(ostream &os, const C &c) {
  bool f = true;
  os << "[";
  for (const auto &x : c) {
    if (!f) os << ", ";
    f = false;
    os << x;
  }
  return os << "]";
}
template <typename T>
void debug(string s, T x) {
  cerr << s << " = " << x << "\n";
}
template <typename T, typename... Args>
void debug(string s, T x, Args... args) {
  cerr << s.substr(0, s.find(',')) << " = " << x << " | ";
  debug(s.substr(s.find(',') + 2), args...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long x;
  cin >> x;
  vector<pair<long long, long long>> v;
  for (long long n = 1;; n++) {
    long long m = (6 * x + n * n * n - n) / (3 * n * n + 3 * n);
    if (n > m) break;
    if ((6 * x + n * n * n - n) % (3 * n * n + 3 * n) == 0)
      v.emplace_back(n, m);
  }
  int sz = v.size();
  for (int i = sz - 1; i >= 0; i--)
    if (v[i].first != v[i].second) v.emplace_back(v[i].second, v[i].first);
  cout << v.size() << "\n";
  for (auto &p : v) cout << p.first << " " << p.second << "\n";
  return 0;
}