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 main() {
  uint64_t n;
  cin >> n;
  vector<pair<uint64_t, uint64_t>> sol;
  n *= 6;
  for (uint64_t i = 1; true; ++i) {
    uint64_t divisor = i * (i + 1);
    if (divisor > n) break;
    uint64_t quotient = n / divisor;
    uint64_t remainder = n % divisor;
    if (quotient < i * 2) break;
    if (remainder != 0) continue;
    uint64_t xm = quotient + i - 1;
    uint64_t mquotient = xm / 3;
    uint64_t mremainder = xm % 3;
    if (mquotient < i) break;
    if (mremainder != 0) continue;
    sol.push_back({i, mquotient});
  }
  if (sol.back().first == sol.back().second)
    cout << sol.size() * 2 - 1 << '\n';
  else
    cout << sol.size() * 2 << '\n';
  cout << sol[0].first << ' ' << sol[0].second;
  for (int i = 1; i < sol.size(); ++i) {
    cout << '\n' << sol[i].first << ' ' << sol[i].second;
  }
  if (sol.back().first != sol.back().second) {
    cout << '\n' << sol.back().second << ' ' << sol.back().first;
  }
  for (int i = sol.size() - 2; i >= 0; --i) {
    cout << '\n' << sol[i].second << ' ' << sol[i].first;
  }
  return 0;
}