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;
using ull = unsigned long long;
using ld = long double;
const long double lot = 2e18;
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  ull x;
  cin >> x;
  vector<pair<ull, ull> > ans;
  ull cap = 6LL * x;
  ull i;
  for (i = 1; i * (i + 1) * (2 * i + 1) < cap; i++) {
    if (cap % (i * (i + 1)) == 0)
      if ((cap / (i * (i + 1)) + i - 1) % 3 == 0) {
        ull temp = (cap / (i * (i + 1)) + i - 1) / 3;
        if (temp > i) ans.push_back(make_pair(i, temp));
      }
  }
  int fix = 0;
  if (i * (i + 1) * (2 * i + 1) == cap) fix = 1;
  cout << 2 * ans.size() + fix << endl;
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << " " << ans[i].second << endl;
  if (fix) cout << i << " " << i << endl;
  for (int i = ans.size() - 1; i >= 0; i--)
    cout << ans[i].second << " " << ans[i].first << endl;
  return 0;
}