#include <bits/stdc++.h>
using namespace std;
set<pair<long long, long long> > s;
set<pair<long long, long long> >::iterator it;
int main() {
  long long xx;
  scanf("%I64d", &xx);
  xx *= 6;
  long long m = 0;
  while (++m < 2000000) {
    long long x = xx;
    if (x % m || x % (m + 1)) continue;
    x /= m, x /= m + 1;
    x += m - 1;
    if (x % 3) continue;
    long long n = x / 3;
    if (n < m) break;
    s.insert(make_pair(m, n));
    s.insert(make_pair(n, m));
  }
  printf("%d\n", s.size());
  for (it = s.begin(); it != s.end(); it++)
    printf("%I64d %I64d\n", it->first, it->second);
}

### Prompt

Your task is to create 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;
set<pair<long long, long long> > s;
set<pair<long long, long long> >::iterator it;
int main() {
  long long xx;
  scanf("%I64d", &xx);
  xx *= 6;
  long long m = 0;
  while (++m < 2000000) {
    long long x = xx;
    if (x % m || x % (m + 1)) continue;
    x /= m, x /= m + 1;
    x += m - 1;
    if (x % 3) continue;
    long long n = x / 3;
    if (n < m) break;
    s.insert(make_pair(m, n));
    s.insert(make_pair(n, m));
  }
  printf("%d\n", s.size());
  for (it = s.begin(); it != s.end(); it++)
    printf("%I64d %I64d\n", it->first, it->second);
}

```