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;
bool compare(pair<unsigned long long, unsigned long long> a,
             pair<unsigned long long, unsigned long long> b) {
  if (a.first < b.first) {
    return true;
  } else if (a.first > b.first) {
    return false;
  } else {
    return (a.second <= b.second);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  unsigned long long x;
  unsigned long long n;
  cin >> x;
  int limit = floor(pow((double)6 * x, (double)1 / 3));
  vector<pair<unsigned long long, unsigned long long> > vec;
  for (unsigned long long m = 1; m <= limit; m++) {
    if ((6 * x) % (m * (m + 1)) == 0) {
      n = (6 * x) / (m * (m + 1)) + m - 1;
      if (n % 3 == 0) {
        n /= 3;
        if (n >= m) {
          vec.push_back(make_pair(m, n));
        }
      }
    }
  }
  int y = vec.size();
  for (int i = 0; i < y; i++) {
    if (vec.at(i).first != vec.at(i).second) {
      vec.push_back(make_pair(vec.at(i).second, vec.at(i).first));
    }
  }
  sort(vec.begin(), vec.end(), compare);
  cout << vec.size() << endl;
  for (int i = 0; i < vec.size(); i++) {
    cout << vec.at(i).first << " " << vec.at(i).second << endl;
  }
  return 0;
}