#include <bits/stdc++.h>
using namespace std;
long long n[1000005], m[1000005];
int main() {
  int k = 0;
  long long x;
  while (scanf("%lld", &x) != EOF) {
    long long c1, c2;
    for (long long i = 1; i <= 10000000; i++) {
      c1 = 6 * x + i * i * i - i;
      c2 = 3 * i * i + 3 * i;
      if (c1 % c2 != 0) continue;
      c1 = c1 / c2;
      if (c1 <= i) {
        if (c1 == i) {
          n[k] = i;
          m[k] = c1;
          k++;
        }
        break;
      }
      n[k] = i;
      m[k] = c1;
      k++;
    }
    if (m[k - 1] != n[k - 1])
      printf("%d\n", 2 * k);
    else
      printf("%d\n", 2 * k - 1);
    for (int i = 0; i < k; i++) printf("%lld %lld\n", n[i], m[i]);
    if (m[k - 1] == n[k - 1]) k--;
    for (int i = k - 1; i >= 0; i--) printf("%lld %lld\n", m[i], n[i]);
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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[1000005], m[1000005];
int main() {
  int k = 0;
  long long x;
  while (scanf("%lld", &x) != EOF) {
    long long c1, c2;
    for (long long i = 1; i <= 10000000; i++) {
      c1 = 6 * x + i * i * i - i;
      c2 = 3 * i * i + 3 * i;
      if (c1 % c2 != 0) continue;
      c1 = c1 / c2;
      if (c1 <= i) {
        if (c1 == i) {
          n[k] = i;
          m[k] = c1;
          k++;
        }
        break;
      }
      n[k] = i;
      m[k] = c1;
      k++;
    }
    if (m[k - 1] != n[k - 1])
      printf("%d\n", 2 * k);
    else
      printf("%d\n", 2 * k - 1);
    for (int i = 0; i < k; i++) printf("%lld %lld\n", n[i], m[i]);
    if (m[k - 1] == n[k - 1]) k--;
    for (int i = k - 1; i >= 0; i--) printf("%lld %lld\n", m[i], n[i]);
  }
  return 0;
}

```