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;
unsigned long long int f(unsigned long long int n, unsigned long long int x) {
  if ((6 * x + n * n * n - n) % (3 * n * n + 3 * n)) return 0;
  return (6 * x + n * n * n - n) / (3 * n * n + 3 * n);
}
unsigned long long int fx(unsigned long long int n, unsigned long long int m) {
  return (n * (n * m * 3 - n * n + m * 3 + 1)) / 6;
}
vector<pair<unsigned long long int, unsigned long long int> > ans;
int main() {
  unsigned long long int n = 1442249;
  int i, j, k, cs;
  unsigned long long int t;
  ;
  ;
  scanf("%llu", &t);
  for (i = 1; i <= 2000000; i++) {
    unsigned long long int j = f(i, t);
    if (j < i) continue;
    ans.push_back({i, j});
  };
  t = ans.size() * 2;
  for (i = 0; i < ans.size(); i++) {
    if (ans[i].first == ans[i].second) t--;
  }
  cout << t << endl;
  for (i = 0; i < ans.size(); i++)
    cout << ans[i].first << " " << ans[i].second << endl;
  for (i = (int)ans.size() - 1; i >= 0; i--) {
    if (ans[i].first == ans[i].second) continue;
    cout << ans[i].second << " " << ans[i].first << endl;
  }
  return 0;
}