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>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long x;
  cin >> x;
  set<pair<long long, long long> > s;
  for (long long i = 1; i * i * i <= 6 * x; i++) {
    if ((3 * (i * i) + (3 * i)) != 0) {
      if (((6 * x) + (i * i * i) - i) % (3 * (i * i) + (3 * i)) == 0) {
        long long n = ((6 * x) + (i * i * i) - i) / (3 * (i * i) + (3 * i)),
                  m = i;
        if (n >= m) {
          if (m != n)
            s.insert(make_pair(m, n)), s.insert(make_pair(n, m));
          else
            s.insert(make_pair(m, n));
        }
      }
    }
  }
  cout << s.size() << "\n";
  for (set<pair<long long, long long> >::iterator it = s.begin(); it != s.end();
       it++)
    cout << (*it).first << " " << (*it).second << "\n";
}