#include <bits/stdc++.h>
using namespace std;
long long n, a = 1, res = 1, b;
vector<pair<long long, long long> > ans;
int main() {
  cin >> n;
  while (res >= 0) {
    res = n - (a * (2 * a + 1) * (a + 1) / 6), b = (a * (a + 1) / 2);
    if (res >= 0 && res % b == 0) {
      ans.push_back({a, a + res / b});
      if (res) ans.push_back({a + res / b, a});
    }
    a++;
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << ' ' << ans[i].second << '\n';
}

### Prompt

Your task is to create a cpp solution to 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;
long long n, a = 1, res = 1, b;
vector<pair<long long, long long> > ans;
int main() {
  cin >> n;
  while (res >= 0) {
    res = n - (a * (2 * a + 1) * (a + 1) / 6), b = (a * (a + 1) / 2);
    if (res >= 0 && res % b == 0) {
      ans.push_back({a, a + res / b});
      if (res) ans.push_back({a + res / b, a});
    }
    a++;
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << ' ' << ans[i].second << '\n';
}

```