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;
long long func(int n, int m) {
  if (n > m) swap(n, m);
  return (long long)n * (n + 1) * (3 * m - n + 1) / 6;
}
long long func2(long long n, long long del) {
  return (long long)n * (n + 1) * (2 * n + 3 * del + 1) / 6;
}
double func3(double n, double del) {
  return n * (n + 1) * (2 * n + 3 * del + 1) / 6.0;
}
int main() {
  long long first;
  cin >> first;
  int limit = 0;
  while ((long long)limit * limit * limit / 3 <= first) ++limit;
  --limit;
  vector<pair<long long, long long> > V;
  for (int n = (int)(1); n < (int)(limit + 1); ++n) {
    long long lo = first * 6 / n / (n + 1);
    lo -= 2 * n + 1;
    lo /= 3;
    if (lo < 0) continue;
    if (func2(n, lo) == first) {
      V.push_back({n, lo + n});
      if (lo) V.push_back({lo + n, n});
    }
  }
  sort(V.begin(), V.end());
  cout << V.size() << endl;
  for (auto first : V) cout << first.first << " " << first.second << endl;
  return 0;
}