#include <bits/stdc++.h>
using namespace std;
vector<unsigned long long> ans;
int main() {
  unsigned long long x;
  scanf("%llu", &x);
  unsigned long long m = x, flag = 0;
  for (unsigned long long n = 1; n <= 1e7; n++) {
    if ((6 * x + n * n * n - n) % (3 * n * n + 3 * n) == 0)
      m = (6 * x + n * n * n - n) / (3 * n * n + 3 * n);
    else if (m < n)
      break;
    else
      continue;
    if (m < n) break;
    ans.push_back(n), ans.push_back(m);
    if (n == m) {
      flag = 1;
      break;
    }
  }
  printf("%d\n", ans.size() - flag);
  for (int i = 0; i < ans.size(); i += 2)
    printf("%llu %llu\n", ans[i], ans[i + 1]);
  for (int i = ans.size() - 1; i >= 0; i -= 2)
    if (ans[i] != ans[i - 1]) printf("%llu %llu\n", ans[i], ans[i - 1]);
}

### 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;
vector<unsigned long long> ans;
int main() {
  unsigned long long x;
  scanf("%llu", &x);
  unsigned long long m = x, flag = 0;
  for (unsigned long long n = 1; n <= 1e7; n++) {
    if ((6 * x + n * n * n - n) % (3 * n * n + 3 * n) == 0)
      m = (6 * x + n * n * n - n) / (3 * n * n + 3 * n);
    else if (m < n)
      break;
    else
      continue;
    if (m < n) break;
    ans.push_back(n), ans.push_back(m);
    if (n == m) {
      flag = 1;
      break;
    }
  }
  printf("%d\n", ans.size() - flag);
  for (int i = 0; i < ans.size(); i += 2)
    printf("%llu %llu\n", ans[i], ans[i + 1]);
  for (int i = ans.size() - 1; i >= 0; i -= 2)
    if (ans[i] != ans[i - 1]) printf("%llu %llu\n", ans[i], ans[i - 1]);
}

```