#include <bits/stdc++.h>
using namespace std;
long long x;
long long ans[10000][2], tot;
int main() {
  cin >> x;
  for (int i = 1; i <= x; i++) {
    long long t = x + (long long)(i - 1) * i / 2 * i -
                  (long long)(i - 1) * (i) * (2 * i - 1) / 6;
    long long tt = (long long)i * i - (long long)(i - 1) * i / 2;
    long long fuck = t / tt;
    if (fuck < i) break;
    if (fuck * tt == t) {
      tot++;
      ans[tot][0] = i;
      ans[tot][1] = fuck;
    }
  }
  int anss = tot * 2;
  if (ans[tot][0] == ans[tot][1]) anss--;
  cout << anss << endl;
  for (int i = 1; i <= tot; i++) printf("%I64d %I64d\n", ans[i][0], ans[i][1]);
  if (ans[tot][0] == ans[tot][1]) tot--;
  for (int i = tot; i >= 1; i--) printf("%I64d %I64d\n", ans[i][1], ans[i][0]);
}

### 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 x;
long long ans[10000][2], tot;
int main() {
  cin >> x;
  for (int i = 1; i <= x; i++) {
    long long t = x + (long long)(i - 1) * i / 2 * i -
                  (long long)(i - 1) * (i) * (2 * i - 1) / 6;
    long long tt = (long long)i * i - (long long)(i - 1) * i / 2;
    long long fuck = t / tt;
    if (fuck < i) break;
    if (fuck * tt == t) {
      tot++;
      ans[tot][0] = i;
      ans[tot][1] = fuck;
    }
  }
  int anss = tot * 2;
  if (ans[tot][0] == ans[tot][1]) anss--;
  cout << anss << endl;
  for (int i = 1; i <= tot; i++) printf("%I64d %I64d\n", ans[i][0], ans[i][1]);
  if (ans[tot][0] == ans[tot][1]) tot--;
  for (int i = tot; i >= 1; i--) printf("%I64d %I64d\n", ans[i][1], ans[i][0]);
}

```