#include <bits/stdc++.h>
using namespace std;
const int N = 4111;
long long x, ans, dp;
set<pair<long long, long long> > aa;
int main(int argc, char** argv) {
  cin >> x;
  for (long long i = 1; i < 2097153; i++) {
    if ((6 * x + i * i * i - i) % (3 * i * i + 3 * i) == 0) {
      if (i <= (6 * x + i * i * i - i) / (3 * i * i + 3 * i)) {
        aa.insert(make_pair(i, (6 * x + i * i * i - i) / (3 * i * i + 3 * i)));
        aa.insert(make_pair((6 * x + i * i * i - i) / (3 * i * i + 3 * i), i));
      }
    }
  }
  cout << aa.size() << endl;
  for (set<pair<long long, long long> >::iterator i = aa.begin(); i != aa.end();
       i++)
    cout << (*i).first << " " << (*i).second << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 = 4111;
long long x, ans, dp;
set<pair<long long, long long> > aa;
int main(int argc, char** argv) {
  cin >> x;
  for (long long i = 1; i < 2097153; i++) {
    if ((6 * x + i * i * i - i) % (3 * i * i + 3 * i) == 0) {
      if (i <= (6 * x + i * i * i - i) / (3 * i * i + 3 * i)) {
        aa.insert(make_pair(i, (6 * x + i * i * i - i) / (3 * i * i + 3 * i)));
        aa.insert(make_pair((6 * x + i * i * i - i) / (3 * i * i + 3 * i), i));
      }
    }
  }
  cout << aa.size() << endl;
  for (set<pair<long long, long long> >::iterator i = aa.begin(); i != aa.end();
       i++)
    cout << (*i).first << " " << (*i).second << endl;
  return 0;
}

```