#include <bits/stdc++.h>
using namespace std;
long long x;
int main() {
  scanf("%lld", &x);
  vector<long long> a, b;
  for (long long m = 1;; ++m) {
    long long n = (6 * x - m + m * m * m) / (3 * m * (m + 1));
    if (m > n) break;
    if (3 * m * m * n - m * m * m + 3 * m * n + m == 6 * x)
      a.push_back(m), b.push_back(n);
  }
  printf("%d\n", 2 * a.size() - (a.back() == b.back()));
  for (int i = 0; i < a.size(); ++i) printf("%lld %lld\n", a[i], b[i]);
  for (int i = a.size() - 1 - (a.back() == b.back()); i >= 0; --i)
    printf("%lld %lld\n", b[i], a[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 x;
int main() {
  scanf("%lld", &x);
  vector<long long> a, b;
  for (long long m = 1;; ++m) {
    long long n = (6 * x - m + m * m * m) / (3 * m * (m + 1));
    if (m > n) break;
    if (3 * m * m * n - m * m * m + 3 * m * n + m == 6 * x)
      a.push_back(m), b.push_back(n);
  }
  printf("%d\n", 2 * a.size() - (a.back() == b.back()));
  for (int i = 0; i < a.size(); ++i) printf("%lld %lld\n", a[i], b[i]);
  for (int i = a.size() - 1 - (a.back() == b.back()); i >= 0; --i)
    printf("%lld %lld\n", b[i], a[i]);
  return 0;
}

```