#include <bits/stdc++.h>
using namespace std;
unsigned long long F[1005][1005];
pair<long long, long long> res[10000005];
int top;
int main() {
  long long n;
  cin >> n;
  long long tr = 1;
  for (long long i = 1; i * (i + 1) * (2 * i + 1) <= n * 6; ++i) {
    long long m = i * (i + 1) * (2 * i + 1) / 6;
    m = n - m;
    if (m % tr == 0) {
      top++;
      res[top].first = i;
      res[top].second = i + m / tr;
    }
    tr += (i + 1);
  }
  int tmp = top;
  for (int i = 1; i <= top; ++i) {
    ++tmp;
    res[tmp].first = res[i].second;
    res[tmp].second = res[i].first;
  }
  sort(res + 1, res + tmp + 1);
  top = tmp;
  int cnt = 0;
  for (int i = 1; i <= top; ++i)
    if (i == 1 || res[i].first != res[i - 1].first ||
        (res[i].first == res[i - 1].first &&
         res[i].second != res[i - 1].second))
      cnt++;
  cout << cnt << endl;
  for (int i = 1; i <= top; ++i)
    if (i == 1 || res[i].first != res[i - 1].first ||
        (res[i].first == res[i - 1].first &&
         res[i].second != res[i - 1].second))
      cout << res[i].first << " " << res[i].second << endl;
  return 0;
}

### Prompt

Please formulate 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;
unsigned long long F[1005][1005];
pair<long long, long long> res[10000005];
int top;
int main() {
  long long n;
  cin >> n;
  long long tr = 1;
  for (long long i = 1; i * (i + 1) * (2 * i + 1) <= n * 6; ++i) {
    long long m = i * (i + 1) * (2 * i + 1) / 6;
    m = n - m;
    if (m % tr == 0) {
      top++;
      res[top].first = i;
      res[top].second = i + m / tr;
    }
    tr += (i + 1);
  }
  int tmp = top;
  for (int i = 1; i <= top; ++i) {
    ++tmp;
    res[tmp].first = res[i].second;
    res[tmp].second = res[i].first;
  }
  sort(res + 1, res + tmp + 1);
  top = tmp;
  int cnt = 0;
  for (int i = 1; i <= top; ++i)
    if (i == 1 || res[i].first != res[i - 1].first ||
        (res[i].first == res[i - 1].first &&
         res[i].second != res[i - 1].second))
      cnt++;
  cout << cnt << endl;
  for (int i = 1; i <= top; ++i)
    if (i == 1 || res[i].first != res[i - 1].first ||
        (res[i].first == res[i - 1].first &&
         res[i].second != res[i - 1].second))
      cout << res[i].first << " " << res[i].second << endl;
  return 0;
}

```