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;
long long X;
vector<pair<int, long long> > ans;
void solve() {
  cin >> X;
  for (int n = 1; n < 1500000; ++n) {
    long long n_ = X * 6 + 1LL * n * n * n - n;
    long long _d = 3LL * n * (n + 1);
    if (n_ % _d == 0 and 1LL * n <= n_ / _d) ans.push_back({n, n_ / _d});
  }
  int square_cnt = 0;
  for (int i = 0; i < ans.size(); ++i)
    if (ans[i].first == ans[i].second) ++square_cnt;
  cout << ans.size() * 2 - square_cnt << "\n";
  for (int i = 0; i < ans.size(); ++i)
    cout << ans[i].first << " " << ans[i].second << "\n";
  for (int i = ans.size() - 1; i >= 0; --i)
    if (ans[i].first != ans[i].second)
      cout << ans[i].second << " " << ans[i].first << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}