#include <bits/stdc++.h>
using namespace std;
long long int ans[2][3000001];
int main() {
  long long int x, num, n, m, val, len = 0, i;
  cin >> x;
  for (i = 1;; i++) {
    if ((i * i * i) > 6 * x) break;
    n = i;
    if ((6 * x) % (n * (n + 1)) == 0) {
      val = 6 * x / n / (n + 1);
      if ((val - 1 + n) % 3 == 0) {
        m = (val - 1 + n) / 3;
        if (n > m) break;
        ans[0][len] = n;
        ans[1][len++] = m;
      }
    }
  }
  if (ans[0][len - 1] != ans[1][len - 1])
    printf("%I64d\n", 2 * len);
  else
    printf("%I64d\n", 2 * len - 1);
  for (i = 0; i < len; i++) {
    printf("%I64d %I64d\n", ans[0][i], ans[1][i]);
  }
  for (i = len - 1; i >= 0; i--) {
    if (ans[1][i] == ans[0][i]) continue;
    printf("%I64d %I64d\n", ans[1][i], ans[0][i]);
  }
  return 0;
}

### Prompt

Please formulate 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 int ans[2][3000001];
int main() {
  long long int x, num, n, m, val, len = 0, i;
  cin >> x;
  for (i = 1;; i++) {
    if ((i * i * i) > 6 * x) break;
    n = i;
    if ((6 * x) % (n * (n + 1)) == 0) {
      val = 6 * x / n / (n + 1);
      if ((val - 1 + n) % 3 == 0) {
        m = (val - 1 + n) / 3;
        if (n > m) break;
        ans[0][len] = n;
        ans[1][len++] = m;
      }
    }
  }
  if (ans[0][len - 1] != ans[1][len - 1])
    printf("%I64d\n", 2 * len);
  else
    printf("%I64d\n", 2 * len - 1);
  for (i = 0; i < len; i++) {
    printf("%I64d %I64d\n", ans[0][i], ans[1][i]);
  }
  for (i = len - 1; i >= 0; i--) {
    if (ans[1][i] == ans[0][i]) continue;
    printf("%I64d %I64d\n", ans[1][i], ans[0][i]);
  }
  return 0;
}

```