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;
const int M = 6000000;
unsigned long long k;
vector<pair<unsigned long long, unsigned long long> > ans;
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  cin >> k;
  for (unsigned long long n = 1; n < M; n++) {
    unsigned long long numenator =
        k - n * (n - 1) * (2 * n - 1) / 6 + n * (n - 1) * n / 2;
    unsigned long long denominator = n * n - (n - 1) * n / 2;
    if (denominator && numenator % denominator == 0) {
      unsigned long long m = numenator / denominator;
      if (m < n) {
        continue;
      }
      ans.push_back(make_pair(n, m));
      if (n != m) {
        ans.push_back(make_pair(m, n));
      }
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].first << " " << ans[i].second << endl;
  }
  return 0;
}