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 n, m;
int main() {
  unsigned long long x;
  vector<pair<unsigned long long, unsigned long long>> results;
  cin >> x;
  if (x == 1) {
    cout << "1\n1 1";
    return 0;
  }
  for (n = 1;; n += 1) {
    m = ((x + (n * (n - 1) * (n + 1)) / 6) / ((n * (n + 1)) / 2));
    if (n > m) break;
    if (((x + (n * (n - 1) * (n + 1)) / 6) % ((n * (n + 1)) / 2))) continue;
    results.push_back(make_pair(n, m));
  }
  m = results.size();
  x = results.size();
  if (results[m - 1].first == results[m - 1].second) {
    m = (m - 1) * 2 + 1;
  } else
    m *= 2;
  cout << m << '\n';
  for (n = 0; n < x; n += 1)
    cout << results[n].first << " " << results[n].second << '\n';
  if (results[n - 1].first == results[n - 1].second) --x;
  for (n = x - 1; n > 0; n -= 1)
    cout << results[n].second << " " << results[n].first << '\n';
  cout << results[n].second << " " << results[n].first;
  return 0;
}