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() {
  ios_base::sync_with_stdio(false);
  long long N;
  cin >> N;
  long long maxv1 = cbrt(N) * 2 + 1;
  vector<pair<long long, long long> > ret;
  for (long long i = 1; i <= maxv1; ++i) {
    if (i % 2 == 0) {
      if (N * 6 % i != 0) continue;
      long long x1 = N * 6 / i - (i - 1) * (2 * i - 1) + i * (i - 1) * 3;
      long long x2 = (i + 1) * 3;
      if (x1 % x2 == 0) {
        if (x1 / x2 < i) break;
        ret.push_back(make_pair(i, x1 / x2));
      }
    } else {
      if (N * 6 % i != 0) continue;
      long long x1 = N * 6 / i - (i - 1) * (2 * i - 1) + i * (i - 1) * 3;
      long long x2 = (i + 1) * 3;
      if (x1 % x2 == 0) {
        if (x1 / x2 < i) break;
        ret.push_back(make_pair(i, x1 / x2));
      }
    }
  }
  int c1 = 0;
  for (int i = 0; i < ret.size(); ++i)
    if (ret[i].first == ret[i].second) c1 += 1;
  cout << ret.size() * 2 - c1 << "\n";
  for (auto i : ret) cout << i.first << " " << i.second << "\n";
  reverse(ret.begin(), ret.end());
  for (auto i : ret)
    if (i.second != i.first) cout << i.second << " " << i.first << "\n";
  return 0;
}