#include <bits/stdc++.h>
using namespace std;
long long x;
int32_t main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  vector<pair<long long, long long> > v;
  cin >> x;
  long long si = 0, si2 = 0;
  for (long long i = 1; si2 < x; i++) {
    si2 += i * i;
    si += i;
    if (((x - si2) >= 0)) {
      if (((x - si2) % (si)) == 0) {
        v.push_back({((x - si2) / si) + i, i});
        if (i != (((x - si2) / si) + i)) {
          v.push_back({i, ((x - si2) / si) + i});
        }
      }
    }
  }
  cout << v.size() << "\n";
  sort(v.begin(), v.end());
  for (auto pr : v) {
    cout << pr.first << " " << pr.second << "\n";
  }
  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;
long long x;
int32_t main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  vector<pair<long long, long long> > v;
  cin >> x;
  long long si = 0, si2 = 0;
  for (long long i = 1; si2 < x; i++) {
    si2 += i * i;
    si += i;
    if (((x - si2) >= 0)) {
      if (((x - si2) % (si)) == 0) {
        v.push_back({((x - si2) / si) + i, i});
        if (i != (((x - si2) / si) + i)) {
          v.push_back({i, ((x - si2) / si) + i});
        }
      }
    }
  }
  cout << v.size() << "\n";
  sort(v.begin(), v.end());
  for (auto pr : v) {
    cout << pr.first << " " << pr.second << "\n";
  }
  return 0;
}

```