#include <bits/stdc++.h>
using namespace std;
vector<long long> N, M;
unsigned long long x;
unsigned long long min(unsigned long long a, unsigned long long b) {
  return (a < b) ? a : b;
}
int main() {
  cin >> x;
  for (unsigned long long m = 1; m <= min(2000000, x); m++) {
    unsigned long long n;
    n = ((6 * x / ((m + 1) * m)) + m + 1) / 3;
    if (n >= m && m * (m + 1) * (3 * n - m + 1) / 6 == x) {
      N.push_back(n);
      M.push_back(m);
    }
  }
  if (N[N.size() - 1] != M[M.size() - 1])
    cout << 2 * N.size() << "\n";
  else
    cout << 2 * N.size() - 1 << "\n";
  for (int i = 0; i < N.size(); i++) {
    cout << M[i] << " " << N[i] << "\n";
  }
  for (int i = (int)N.size() - 1; i >= 0; i--) {
    if (M[i] != N[i]) cout << N[i] << " " << M[i] << "\n";
  }
  return 0;
}

### Prompt

Generate 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;
vector<long long> N, M;
unsigned long long x;
unsigned long long min(unsigned long long a, unsigned long long b) {
  return (a < b) ? a : b;
}
int main() {
  cin >> x;
  for (unsigned long long m = 1; m <= min(2000000, x); m++) {
    unsigned long long n;
    n = ((6 * x / ((m + 1) * m)) + m + 1) / 3;
    if (n >= m && m * (m + 1) * (3 * n - m + 1) / 6 == x) {
      N.push_back(n);
      M.push_back(m);
    }
  }
  if (N[N.size() - 1] != M[M.size() - 1])
    cout << 2 * N.size() << "\n";
  else
    cout << 2 * N.size() - 1 << "\n";
  for (int i = 0; i < N.size(); i++) {
    cout << M[i] << " " << N[i] << "\n";
  }
  for (int i = (int)N.size() - 1; i >= 0; i--) {
    if (M[i] != N[i]) cout << N[i] << " " << M[i] << "\n";
  }
  return 0;
}

```