#include <bits/stdc++.h>
using namespace std;
long long n, s, t, c, a[99], b[99];
int main() {
  cin >> s;
  s *= 6;
  for (n = 1; n * (n + 1) * (n + n + 1) <= s; ++n) {
    t = s - n * (n + 1) * (n + n + 1);
    if (t % (3 * n * (n + 1)) == 0) a[++c] = n, b[c] = n + t / 3 / n / (n + 1);
  }
  if (!c)
    puts("0");
  else {
    cout << c + c - (a[c] == b[c]);
    for (n = 1; n <= c; n++) cout << endl << a[n] << ' ' << b[n];
    for (n -= a[c] == b[c]; --n;) cout << endl << b[n] << ' ' << a[n];
  }
  return 0;
}

### 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;
long long n, s, t, c, a[99], b[99];
int main() {
  cin >> s;
  s *= 6;
  for (n = 1; n * (n + 1) * (n + n + 1) <= s; ++n) {
    t = s - n * (n + 1) * (n + n + 1);
    if (t % (3 * n * (n + 1)) == 0) a[++c] = n, b[c] = n + t / 3 / n / (n + 1);
  }
  if (!c)
    puts("0");
  else {
    cout << c + c - (a[c] == b[c]);
    for (n = 1; n <= c; n++) cout << endl << a[n] << ' ' << b[n];
    for (n -= a[c] == b[c]; --n;) cout << endl << b[n] << ' ' << a[n];
  }
  return 0;
}

```