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() {
  long long int x;
  cin >> x;
  long long int y = 6 * x;
  long long int equal = -1;
  long long int upperLim = pow(3 * x, 1.0 / 3);
  std::vector<pair<long long int, long long int> > v;
  for (long long int i = 1; i <= upperLim; i++) {
    if (y % (i * (i + 1)) == 0) {
      long long int n = y / (i * (i + 1));
      n = n + i - 1;
      if (n % 3 == 0) {
        n = n / 3;
        if (i == n) {
          equal = i;
          break;
        } else {
          v.push_back(make_pair(i, n));
        }
      }
    }
  }
  long long int l = v.size();
  if (equal != -1) {
    cout << 2 * l + 1 << endl;
    for (int i = 0; i < l; i++) {
      cout << v[i].first << " " << v[i].second << endl;
    }
    cout << equal << " " << equal << endl;
    for (int i = l - 1; i >= 0; i--) {
      cout << v[i].second << " " << v[i].first << endl;
    }
  } else {
    cout << 2 * l << endl;
    for (int i = 0; i < l; i++) {
      cout << v[i].first << " " << v[i].second << endl;
    }
    for (int i = l - 1; i >= 0; i--) {
      cout << v[i].second << " " << v[i].first << endl;
    }
  }
}