#include <bits/stdc++.h>
using namespace std;
vector<pair<long long int, long long int> > v;
int main() {
  long long x;
  cin >> x;
  x *= 6;
  long long prev, s = 0;
  for (long long i = 1;; i++) {
    s += i * i;
    if (x / 6 < s) break;
    if ((x % (i * (i + 1))) == 0) {
      long long p = (x / (i * (i + 1)));
      p += i - 1;
      if (p % 3 == 0) {
        p /= 3;
        v.push_back({i, p});
        prev = p;
      }
    }
  }
  int r = v.size();
  for (int i = 0; i < r; i++) {
    if (v[i].first != v[i].second) v.push_back({v[i].second, v[i].first});
  }
  sort(v.begin(), v.end());
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++)
    printf("%I64d %I64d\n", v[i].first, v[i].second);
}

### Prompt

In CPP, your task is to solve 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 int, long long int> > v;
int main() {
  long long x;
  cin >> x;
  x *= 6;
  long long prev, s = 0;
  for (long long i = 1;; i++) {
    s += i * i;
    if (x / 6 < s) break;
    if ((x % (i * (i + 1))) == 0) {
      long long p = (x / (i * (i + 1)));
      p += i - 1;
      if (p % 3 == 0) {
        p /= 3;
        v.push_back({i, p});
        prev = p;
      }
    }
  }
  int r = v.size();
  for (int i = 0; i < r; i++) {
    if (v[i].first != v[i].second) v.push_back({v[i].second, v[i].first});
  }
  sort(v.begin(), v.end());
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++)
    printf("%I64d %I64d\n", v[i].first, v[i].second);
}

```