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.

<image>

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;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
long long hoge = 1000000000000000000;
int main() {
  long long int x;
  cin >> x;
  long long int n, a, b;
  long long int cnt = 0;
  vector<pair<long long, long long> > ans;
  for (n = 1; n * (n + 1) * (2 * n + 1) / 6 <= x; n++) {
    if ((6 * x) % (n * (n + 1)) == 0) {
      a = 6 * x / (n * (n + 1)) + (long long)n - (long long)1;
      if (a % 3 == 0) {
        cnt++;
        ans.push_back(make_pair((n), (a / 3)));
        if (n != a / 3) {
          ans.push_back(make_pair((a / 3), (n)));
          cnt++;
        }
      }
    }
  }
  sort(ans.begin(), ans.end());
  cout << cnt << endl;
  for (int(i) = 0; (i) < (int)(ans.size()); ++(i))
    cout << ans[i].first << " " << ans[i].second << endl;
}