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 maxn = 1e6 + 2;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, n, m, first, c = 0;
  vector<long long int> a, b;
  cin >> first;
  if (first == 1) {
    cout << 1 << "\n" << 1 << " " << 1;
    return 0;
  }
  for (n = 1; n * n <= first; ++n) {
    long long int o = 6 * first + n * n * n - n;
    long long int d = 3 * n * (n + 1);
    if (n > o / d) break;
    if (o % d == 0) {
      long long int m = o / d;
      a.push_back(n);
      b.push_back(m);
      if (n == m) ++c;
    }
  }
  cout << 2 * a.size() - c << "\n";
  for (i = 0; i < a.size(); ++i) cout << a[i] << " " << b[i] << "\n";
  for (i = (a.size() - 1) ? a.size() - 1 : 0; i >= 0; --i)
    if (b[i] != a[i]) cout << b[i] << " " << a[i] << "\n";
}