#include <bits/stdc++.h>
using namespace std;
long long a[100000];
long long b[100000];
int main() {
  long long x;
  long long m, n;
  cin >> x;
  int pos = 0;
  for (m = 1; m < 1452200; m++) {
    n = ((6 * x) / (m * (m + 1)) + m - 1);
    if (n % 3 != 0) continue;
    n = n / 3;
    if (m > n) break;
    if (6 * x == (m * (m + 1) * (3 * n - m + 1))) {
      a[pos] = m;
      b[pos++] = n;
    }
  }
  long long total = 2 * pos;
  if (a[pos - 1] == b[pos - 1]) {
    total--;
  }
  cout << total << endl;
  for (int i = 0; i < pos; i++) {
    cout << a[i] << " " << b[i] << endl;
  }
  if (a[pos - 1] == b[pos - 1]) {
    pos--;
  }
  for (int i = 0; i < pos; i++) {
    cout << b[pos - i - 1] << " " << a[pos - i - 1] << endl;
  }
}

### Prompt

Create a solution in cpp for 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 a[100000];
long long b[100000];
int main() {
  long long x;
  long long m, n;
  cin >> x;
  int pos = 0;
  for (m = 1; m < 1452200; m++) {
    n = ((6 * x) / (m * (m + 1)) + m - 1);
    if (n % 3 != 0) continue;
    n = n / 3;
    if (m > n) break;
    if (6 * x == (m * (m + 1) * (3 * n - m + 1))) {
      a[pos] = m;
      b[pos++] = n;
    }
  }
  long long total = 2 * pos;
  if (a[pos - 1] == b[pos - 1]) {
    total--;
  }
  cout << total << endl;
  for (int i = 0; i < pos; i++) {
    cout << a[i] << " " << b[i] << endl;
  }
  if (a[pos - 1] == b[pos - 1]) {
    pos--;
  }
  for (int i = 0; i < pos; i++) {
    cout << b[pos - i - 1] << " " << a[pos - i - 1] << endl;
  }
}

```