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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int inf = 0x3f3f3f3f;
long long infl = 0x3f3f3f3f3f3f3f3fLL;
long double infd = 1.0 / 0.0;
const long long MOD = 1e9 + 7;
const long double pi = 2 * acos(0.0);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long x;
  cin >> x;
  set<pair<long long, long long>> st;
  long long lim = (long long)cbrt(7 * x);
  for (long long n = 1; n <= lim; n++) {
    long long z = 6 * x + n * n * n - n;
    long long p = 3 * n * (n + 1);
    if (z % p == 0 && n <= z / p) {
      st.insert({n, z / p});
      st.insert({z / p, n});
    }
  }
  cout << (int)st.size() << '\n';
  for (auto i : st) cout << i.first << ' ' << i.second << '\n';
  return 0;
}