#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000000;100000000000")
const unsigned long long int NMAX = 10000000;
const double eps = 0.0001;
using namespace std;
unsigned long long int m(unsigned long long int x, unsigned long long int n) {
  unsigned long long int ans = (6 * x / n - 1 + n * n) / 3 / (n + 1);
  if ((3 * n * n * ans + 3 * n * ans - n * n * n + n) != x * 6)
    return 0;
  else
    return ans;
}
int main() {
  ios::sync_with_stdio(0);
  unsigned long long int x;
  cin >> x;
  unsigned long long int counter = 0;
  vector<pair<unsigned long long int, unsigned long long int> > ans;
  unsigned long long int t;
  unsigned long long int f =
      (unsigned long long int)min(2 * (pow(x, 1.0 / 3.0)), sqrt(x));
  for (unsigned long long int i = 1; i <= f; i++) {
    t = m(x, i);
    if (t >= i && t != 0) {
      counter++;
      ans.push_back(make_pair(i, t));
    }
  }
  if (ans[counter - 1].first != ans[counter - 1].second)
    cout << 2 * counter << endl;
  else
    cout << 2 * counter - 1 << endl;
  for (unsigned long long int i = 0; i < counter; i++)
    cout << ans[i].first << " " << ans[i].second << endl;
  if (ans[counter - 1].first != ans[counter - 1].second)
    cout << ans[counter - 1].second << " " << ans[counter - 1].first << endl;
  for (unsigned long long int i = 1; i < counter; i++)
    cout << ans[counter - 1 - i].second << " " << ans[counter - 1 - i].first
         << endl;
  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>
#pragma comment(linker, "/STACK:100000000000;100000000000")
const unsigned long long int NMAX = 10000000;
const double eps = 0.0001;
using namespace std;
unsigned long long int m(unsigned long long int x, unsigned long long int n) {
  unsigned long long int ans = (6 * x / n - 1 + n * n) / 3 / (n + 1);
  if ((3 * n * n * ans + 3 * n * ans - n * n * n + n) != x * 6)
    return 0;
  else
    return ans;
}
int main() {
  ios::sync_with_stdio(0);
  unsigned long long int x;
  cin >> x;
  unsigned long long int counter = 0;
  vector<pair<unsigned long long int, unsigned long long int> > ans;
  unsigned long long int t;
  unsigned long long int f =
      (unsigned long long int)min(2 * (pow(x, 1.0 / 3.0)), sqrt(x));
  for (unsigned long long int i = 1; i <= f; i++) {
    t = m(x, i);
    if (t >= i && t != 0) {
      counter++;
      ans.push_back(make_pair(i, t));
    }
  }
  if (ans[counter - 1].first != ans[counter - 1].second)
    cout << 2 * counter << endl;
  else
    cout << 2 * counter - 1 << endl;
  for (unsigned long long int i = 0; i < counter; i++)
    cout << ans[i].first << " " << ans[i].second << endl;
  if (ans[counter - 1].first != ans[counter - 1].second)
    cout << ans[counter - 1].second << " " << ans[counter - 1].first << endl;
  for (unsigned long long int i = 1; i < counter; i++)
    cout << ans[counter - 1 - i].second << " " << ans[counter - 1 - i].first
         << endl;
  return 0;
}

```