#include <bits/stdc++.h>
using namespace std;
int main() {
  long long i, x, a[100000], b[100000], h = 0, n;
  long long m, f, c = 0;
  scanf("%lld", &x);
  if (x == 1) {
    printf("1\n1 1\n");
  } else {
    for (i = 1; i <= 10000000; i++) {
      m = (6 * x + i * (i - 1) * (i + 1));
      f = (3 * i * (i + 1));
      if (m % f == 0 && i <= m / f) {
        a[h] = i;
        b[h++] = m / f;
      }
    }
    for (i = 0; i < h; i++)
      if (a[i] == b[i]) c++;
    printf("%lld\n", 2 * h - c);
    for (i = 0; i < h; i++) printf("%lld %lld\n", a[i], b[i]);
    for (i = h - 1; i >= 0; i--)
      if (a[i] != b[i]) printf("%lld %lld\n", b[i], a[i]);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int main() {
  long long i, x, a[100000], b[100000], h = 0, n;
  long long m, f, c = 0;
  scanf("%lld", &x);
  if (x == 1) {
    printf("1\n1 1\n");
  } else {
    for (i = 1; i <= 10000000; i++) {
      m = (6 * x + i * (i - 1) * (i + 1));
      f = (3 * i * (i + 1));
      if (m % f == 0 && i <= m / f) {
        a[h] = i;
        b[h++] = m / f;
      }
    }
    for (i = 0; i < h; i++)
      if (a[i] == b[i]) c++;
    printf("%lld\n", 2 * h - c);
    for (i = 0; i < h; i++) printf("%lld %lld\n", a[i], b[i]);
    for (i = h - 1; i >= 0; i--)
      if (a[i] != b[i]) printf("%lld %lld\n", b[i], a[i]);
  }
  return 0;
}

```