#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x;
  scanf("%lld", &x);
  long long b, l;
  set<pair<long long, long long> > ans;
  for (b = 1; (long long)b * b <= 3 * x / b + b - 1; b++) {
    long long y = 3 * x;
    long long z = b * (b + 1) / 2;
    if (y % z != 0) continue;
    y /= z;
    y = y - 1 + b;
    if (y > 0 && y % 3 == 0) {
      ans.insert(make_pair(b, y / 3));
      if (b != y / 3) {
        ans.insert(make_pair(y / 3, b));
      }
    }
  }
  printf("%lu\n", ans.size());
  set<pair<long long, long long> >::iterator it = ans.begin();
  ;
  while (it != ans.end()) {
    pair<long long, long long> c = *it;
    printf("%lld %lld\n", c.first, c.second);
    it++;
  }
  return 0;
}

### 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;
int main() {
  long long x;
  scanf("%lld", &x);
  long long b, l;
  set<pair<long long, long long> > ans;
  for (b = 1; (long long)b * b <= 3 * x / b + b - 1; b++) {
    long long y = 3 * x;
    long long z = b * (b + 1) / 2;
    if (y % z != 0) continue;
    y /= z;
    y = y - 1 + b;
    if (y > 0 && y % 3 == 0) {
      ans.insert(make_pair(b, y / 3));
      if (b != y / 3) {
        ans.insert(make_pair(y / 3, b));
      }
    }
  }
  printf("%lu\n", ans.size());
  set<pair<long long, long long> >::iterator it = ans.begin();
  ;
  while (it != ans.end()) {
    pair<long long, long long> c = *it;
    printf("%lld %lld\n", c.first, c.second);
    it++;
  }
  return 0;
}

```