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;
unsigned long long x;
stack<pair<unsigned long long, unsigned long long> > q;
vector<pair<unsigned long long, unsigned long long> > sol;
int main() {
  cin >> x;
  x *= 6;
  for (unsigned long long n = 1;; ++n) {
    unsigned long long k = 2 * n * n * n;
    if (k > x) break;
    k += 3 * n * n;
    if (k > x) break;
    k += n;
    if (k > x) break;
    unsigned long long w = 3 * n * (n + 1);
    if ((x - k) % w == 0) {
      unsigned long long m = (x - k) / w + n;
      sol.push_back(make_pair(n, m));
      q.push(make_pair(m, n));
    }
  }
  while (!q.empty()) {
    if (q.top().first != q.top().second)
      sol.push_back(make_pair(q.top().first, q.top().second));
    q.pop();
  }
  cout << sol.size() << endl;
  for (int i = 0; i < sol.size(); ++i) {
    cout << sol[i].first << ' ' << sol[i].second << endl;
  }
  return 0;
}