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;
long long calc_m(long long n, long long x) {
  long long ret = -1;
  if ((6 * x - n + n * n * n) % (3 * n * (n + 1)) == 0) {
    ret = (6 * x - n + n * n * n) / (3 * n * (n + 1));
  }
  return ret;
}
int main() {
  long long x;
  cin >> x;
  long long n = 1;
  vector<pair<long long, long long> > ans;
  if (x == 1)
    ans.push_back(make_pair(1, 1));
  else {
    while (6 * x >= 2 * n * n * n + 3 * n * n + n) {
      long long m = calc_m(n, x);
      if (m != -1) {
        ans.push_back(make_pair(n, m));
      }
      n++;
    }
    long long lim = ans.size();
    if (ans[ans.size() - 1].first == ans[ans.size() - 1].second) lim--;
    for (long long i = lim - 1; i >= 0; --i) {
      ans.push_back(make_pair(ans[i].second, ans[i].first));
    }
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < ans.size(); ++i) {
    cout << ans[i].first << " " << ans[i].second << endl;
  }
}