#include <bits/stdc++.h>
using namespace std;
int main() {
  long long i, n;
  cin >> n;
  vector<pair<long long, long long> > res;
  for (i = 1; i < 2 * 1e6; i++) {
    if ((6 * n) % (i * (i + 1)) == 0) {
      long long d = (6 * n) / (i * (i + 1)) - (2 * i + 1);
      if (d >= 0 && d % 3 == 0) {
        d /= 3;
        res.push_back({i, i + d});
        if (d != 0) res.push_back({i + d, i});
      }
    }
  }
  auto comp = [](const pair<long long, long long>& a,
                 const pair<long long, long long>& b) {
    return a.first == b.first ? a.second < b.second : a.first < b.first;
  };
  sort(res.begin(), res.end(), comp);
  cout << res.size() << endl;
  for (pair<long long, long long> p : res)
    cout << p.first << " " << p.second << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve the following 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.

<image>

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long i, n;
  cin >> n;
  vector<pair<long long, long long> > res;
  for (i = 1; i < 2 * 1e6; i++) {
    if ((6 * n) % (i * (i + 1)) == 0) {
      long long d = (6 * n) / (i * (i + 1)) - (2 * i + 1);
      if (d >= 0 && d % 3 == 0) {
        d /= 3;
        res.push_back({i, i + d});
        if (d != 0) res.push_back({i + d, i});
      }
    }
  }
  auto comp = [](const pair<long long, long long>& a,
                 const pair<long long, long long>& b) {
    return a.first == b.first ? a.second < b.second : a.first < b.first;
  };
  sort(res.begin(), res.end(), comp);
  cout << res.size() << endl;
  for (pair<long long, long long> p : res)
    cout << p.first << " " << p.second << endl;
  return 0;
}

```