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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  vector<pair<unsigned long long, unsigned long long> > v;
  unsigned long long target;
  cin >> target;
  unsigned long long base_l = 1, base_r = target + 1;
  for (unsigned long long n = 1; n * (n + 1) * (2 * n + 1) <= 6 * target; n++) {
    unsigned long long l = base_l, r = base_r;
    unsigned long long base_n1 = n * (n + 1) * (2 * n + 1),
                       base_n2 = 3 * n * (n + 1);
    while (r - l > 1) {
      unsigned long long mid = (r + l) / 2;
      if (base_n1 + (mid - n) * base_n2 <= 6 * target)
        l = mid;
      else
        r = mid;
    }
    base_r = r;
    if (base_n1 + (l - n) * base_n2 == 6 * target) v.push_back({n, l});
  }
  cout << 2 * v.size() - (v.back().first != v.back().second ? 0 : 1) << '\n';
  for (auto p : v) cout << p.first << " " << p.second << '\n';
  for (int i = v.size() - 1; i >= 0; i--) {
    auto p = v[i];
    unsigned long long a = p.first, b = p.second;
    if (a != b) cout << b << " " << a << '\n';
  }
  return 0;
}