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 mod = 1e9 + 7, maxn = 100005, ln = 18;
const int inf = int(1e9), shift = 30, etf = mod - 1;
vector<pair<unsigned long long int, unsigned long long int> > vec;
int main() {
  unsigned long long int x, n, m, p = 1, q = 2000000, mid, cal, ans = 0;
  cin >> x;
  while (p != q) {
    mid = (p + q) >> 1;
    cal = (mid * (mid + 1) * (2 * mid + 1)) / 6;
    if (cal < x)
      p = mid + 1;
    else
      q = mid;
  }
  cal = (p * (p + 1) * (2 * p + 1)) / 6;
  if (cal >= x) p--;
  if (cal == x) ans++;
  for (int i = 1; i < p + 1; i++) {
    q = i;
    if ((6 * x) % (q * (q + 1)) == 0) {
      mid = (((6 * x) / (q * (q + 1))) + q - 1);
      if (mid % 3 == 0) {
        mid /= 3;
        vec.push_back(
            pair<unsigned long long int, unsigned long long int>(mid, i));
      }
    }
  }
  ans += 2 * (int)vec.size();
  cout << ans << endl;
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i].second << " " << vec[i].first << endl;
  }
  if (cal == x) cout << p + 1 << " " << p + 1 << endl;
  for (int i = vec.size() - 1; i >= 0; i--) {
    cout << vec[i].first << " " << vec[i].second << endl;
  }
  return 0;
}