#include <bits/stdc++.h>
using namespace std;
long long check(long long a, long long b) {
  long long ret = (a * (a + 1)) * (3 * b - a + 1);
  ret /= 6;
  return ret;
}
int main() {
  long long x;
  cin >> x;
  set<pair<long long, long long> > st;
  for (long long i = 1; i <= x; i++) {
    long long n = i;
    long long m = 6 * x;
    m /= n;
    m /= (n + 1);
    m += n;
    m -= 1;
    m /= 3;
    if (m >= n && m > 0 && n > 0) {
      if (check(n, m) != x) continue;
      st.insert(make_pair(n, m));
      st.insert(make_pair(m, n));
    } else
      break;
  }
  cout << st.size() << endl;
  for (set<pair<long long, long long> >::iterator it = st.begin();
       it != st.end(); it++)
    cout << (*it).first << " " << (*it).second << endl;
}

### Prompt

Generate 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 check(long long a, long long b) {
  long long ret = (a * (a + 1)) * (3 * b - a + 1);
  ret /= 6;
  return ret;
}
int main() {
  long long x;
  cin >> x;
  set<pair<long long, long long> > st;
  for (long long i = 1; i <= x; i++) {
    long long n = i;
    long long m = 6 * x;
    m /= n;
    m /= (n + 1);
    m += n;
    m -= 1;
    m /= 3;
    if (m >= n && m > 0 && n > 0) {
      if (check(n, m) != x) continue;
      st.insert(make_pair(n, m));
      st.insert(make_pair(m, n));
    } else
      break;
  }
  cout << st.size() << endl;
  for (set<pair<long long, long long> >::iterator it = st.begin();
       it != st.end(); it++)
    cout << (*it).first << " " << (*it).second << endl;
}

```