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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long x;
long long k;
const int maxn = 1442250;
vector<pair<long long, long long> > ans;
int main() {
  cin >> x;
  for (long long a = 1; a <= maxn && a <= x; a++) {
    long long b = (6 * x + (long long)a * a * a + a) / 3 / a / (a + 1);
    long long tmp = (3 * a * b + 3 * a * a * b - a * a * a + a) / 6;
    if (tmp == x && a <= b) {
      ans.push_back(make_pair(a, b));
      if (a != b) ans.push_back(make_pair(b, a));
    }
  }
  sort(ans.begin(), ans.end(),
       [](pair<long long, long long> a, pair<long long, long long> b) {
         if (a.first == b.first)
           return a.second < b.second;
         else
           return a.first < b.first;
       });
  cout << ans.size() << endl;
  for (auto i = ans.begin(); i != ans.end(); i++)
    cout << (*i).first << " " << (*i).second << endl;
  return 0;
}