#include <bits/stdc++.h>
using namespace std;
long long c[205000];
long long d[205000];
int main() {
  long long n, m, k, l, i, j, r;
  while (~scanf("%I64d", &n)) {
    r = 0;
    l = 1;
    for (i = 1; i <= 10000000; i++) {
      if ((6 * n + i * i * i - i) % (3 * i * i + 3 * i) == 0) {
        long long k = (6 * n + i * i * i - i) / (3 * i * i + 3 * i);
        if (k >= i) {
          if (k == i) r++;
          c[l] = i;
          d[l] = k;
          l++;
        }
      }
    }
    printf("%I64d\n", ((l - 1) * 2 - r));
    for (i = 1; i < l; i++) printf("%I64d %I64d\n", c[i], d[i]);
    for (i = l - 1; i >= 1; i--) {
      if (d[i] == c[i]) continue;
      printf("%I64d %I64d\n", d[i], c[i]);
    }
  }
  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 c[205000];
long long d[205000];
int main() {
  long long n, m, k, l, i, j, r;
  while (~scanf("%I64d", &n)) {
    r = 0;
    l = 1;
    for (i = 1; i <= 10000000; i++) {
      if ((6 * n + i * i * i - i) % (3 * i * i + 3 * i) == 0) {
        long long k = (6 * n + i * i * i - i) / (3 * i * i + 3 * i);
        if (k >= i) {
          if (k == i) r++;
          c[l] = i;
          d[l] = k;
          l++;
        }
      }
    }
    printf("%I64d\n", ((l - 1) * 2 - r));
    for (i = 1; i < l; i++) printf("%I64d %I64d\n", c[i], d[i]);
    for (i = l - 1; i >= 1; i--) {
      if (d[i] == c[i]) continue;
      printf("%I64d %I64d\n", d[i], c[i]);
    }
  }
  return 0;
}

```