#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<pair<long long int, long long int> > v;
  long long int x;
  cin >> x;
  long long int d;
  long long int u;
  for (long long int n = 1; n <= min(x, 5000000LL); n++) {
    d = n * n - ((n - 1) * n) / 2;
    u = x + (n * n * (n - 1)) / 2 - (n * (n - 1) * (2 * n - 1)) / 6;
    if (u % d == 0) {
      u /= d;
      if (u >= n) {
        v.push_back({n, u});
        if (u != n) v.push_back({u, n});
      }
    }
  }
  cout << v.size() << "\n";
  sort(v.begin(), v.end());
  for (auto x : v) {
    cout << x.first << " " << x.second << "\n";
  }
}

### Prompt

Your challenge is to write 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;
int main() {
  vector<pair<long long int, long long int> > v;
  long long int x;
  cin >> x;
  long long int d;
  long long int u;
  for (long long int n = 1; n <= min(x, 5000000LL); n++) {
    d = n * n - ((n - 1) * n) / 2;
    u = x + (n * n * (n - 1)) / 2 - (n * (n - 1) * (2 * n - 1)) / 6;
    if (u % d == 0) {
      u /= d;
      if (u >= n) {
        v.push_back({n, u});
        if (u != n) v.push_back({u, n});
      }
    }
  }
  cout << v.size() << "\n";
  sort(v.begin(), v.end());
  for (auto x : v) {
    cout << x.first << " " << x.second << "\n";
  }
}

```