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;
int dr[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dc[] = {0, -1, 0, 1, -1, 1, -1, 1};
const int MAX = 1e6 + 7;
const int MOD = 1e7 + 7;
vector<pair<long long, long long> > ans;
int main() {
  long long x;
  cin >> x;
  for (long long n = 1, m; n * n * n <= 8 * x; ++n) {
    long long a = 6 * x + n * n * n - n, b = 3 * (n * n + n);
    if (a % b != 0) continue;
    m = a / b;
    if (m < n) break;
    ans.push_back(pair<long long, long long>(n, m));
  }
  cout << 2 * (long long)ans.size() - (ans[(long long)ans.size() - 1].first ==
                                       ans[(long long)ans.size() - 1].second)
       << "\n";
  for (int i = 0; i < (long long)ans.size(); ++i)
    cout << ans[i].first << " " << ans[i].second << "\n";
  for (int i = (long long)ans.size() - 1; i >= 0; --i)
    if (ans[i].first != ans[i].second)
      cout << ans[i].second << " " << ans[i].first << "\n";
}