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() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cout.tie(0);
  unsigned long long n, k = 1, r, a, b, c = 1e10;
  vector<pair<unsigned long long, unsigned long long>> v;
  cin >> n;
  n *= 6;
  r = floor(pow(n, 1.0 / 3.0));
  while (k <= min(r, c)) {
    if (n % (k * (k + 1)) == 0) {
      a = n / (k * (k + 1));
      b = a + 2 + k;
      if (k <= b / 3 - 1 && b % 3 == 0) {
        v.push_back({k, b / 3 - 1});
        c = b / 3 - 1;
      }
    }
    k++;
  }
  if (v[v.size() - 1].first == v[v.size() - 1].second)
    cout << 2 * v.size() - 1 << endl;
  else
    cout << 2 * v.size() << endl;
  for (int i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << "\n";
  if (v[v.size() - 1].first != v[v.size() - 1].second)
    cout << v[v.size() - 1].second << " " << v[v.size() - 1].first << "\n";
  for (int i = v.size() - 2; i >= 0; i--)
    cout << v[i].second << " " << v[i].first << "\n";
  return 0;
}