#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
vector<pair<long long, long long> > ans;
int main() {
  long long x;
  cin >> x;
  x *= 6;
  for (long long b = 1; b * b * b <= x; ++b) {
    if (x % b == 0 && x % (b + 1) == 0) {
      long long k = x / b / (b + 1) - (2 * b + 1);
      if (k < 0) break;
      if (k % 3 == 0) ans.push_back(make_pair(b, b + k / 3));
    }
  }
  int n = ans.size();
  for (int i = n - 1; i >= 0; --i)
    if (ans[i].first < ans[i].second)
      ans.push_back(make_pair(ans[i].second, ans[i].first));
  printf("%d\n", (int)ans.size());
  for (auto pair : ans) printf("%I64d %I64d\n", pair.first, pair.second);
  return 0;
}

### 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;
const int N = 2e6 + 10;
vector<pair<long long, long long> > ans;
int main() {
  long long x;
  cin >> x;
  x *= 6;
  for (long long b = 1; b * b * b <= x; ++b) {
    if (x % b == 0 && x % (b + 1) == 0) {
      long long k = x / b / (b + 1) - (2 * b + 1);
      if (k < 0) break;
      if (k % 3 == 0) ans.push_back(make_pair(b, b + k / 3));
    }
  }
  int n = ans.size();
  for (int i = n - 1; i >= 0; --i)
    if (ans[i].first < ans[i].second)
      ans.push_back(make_pair(ans[i].second, ans[i].first));
  printf("%d\n", (int)ans.size());
  for (auto pair : ans) printf("%I64d %I64d\n", pair.first, pair.second);
  return 0;
}

```