#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > v;
long long x;
int main() {
  long long x;
  cin >> x;
  long long cons = x * 6;
  long long cntr = 0;
  for (long long i = 1; i <= 10000000; i++) {
    if (cons % (i * (i + 1)) == 0LL) {
      long long t = cons / (i * (i + 1));
      long long x = t + i - 1;
      if (x % 3 != 0LL) continue;
      x = x / 3;
      if (x >= i) {
        cntr++;
        v.push_back(make_pair(x, i));
        if (x != i) v.push_back(make_pair(i, x));
      }
    }
  }
  sort(v.begin(), v.end());
  int sz = v.size();
  cout << sz << endl;
  for (int i = 0; i < sz; i++) cout << v[i].first << " " << v[i].second << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
vector<pair<long long, long long> > v;
long long x;
int main() {
  long long x;
  cin >> x;
  long long cons = x * 6;
  long long cntr = 0;
  for (long long i = 1; i <= 10000000; i++) {
    if (cons % (i * (i + 1)) == 0LL) {
      long long t = cons / (i * (i + 1));
      long long x = t + i - 1;
      if (x % 3 != 0LL) continue;
      x = x / 3;
      if (x >= i) {
        cntr++;
        v.push_back(make_pair(x, i));
        if (x != i) v.push_back(make_pair(i, x));
      }
    }
  }
  sort(v.begin(), v.end());
  int sz = v.size();
  cout << sz << endl;
  for (int i = 0; i < sz; i++) cout << v[i].first << " " << v[i].second << endl;
  return 0;
}

```