#include <bits/stdc++.h>
using namespace std;
struct thing {
  long long x, y;
} ans[1000000];
long long a, n, tot, same;
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  long long i = 1;
  while (true) {
    long long b = (i - 1) * i * (2 * i - 1) / 6 - i * i * (i - 1) / 2;
    if (b >= n) break;
    long long k = i * (i + 1) / 2;
    if ((n - b) / k < i) break;
    if ((n - b) % k == 0) {
      ans[++tot].x = i;
      ans[tot].y = (n - b) / k;
      if (ans[tot].x == ans[tot].y) same++;
    }
    i++;
  }
  cout << 2 * tot - same << endl;
  for (int i = 1; i <= tot; i++) cout << ans[i].x << " " << ans[i].y << endl;
  for (int i = tot; i; i--)
    if (ans[i].x != ans[i].y) cout << ans[i].y << " " << ans[i].x << endl;
}

### 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;
struct thing {
  long long x, y;
} ans[1000000];
long long a, n, tot, same;
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  long long i = 1;
  while (true) {
    long long b = (i - 1) * i * (2 * i - 1) / 6 - i * i * (i - 1) / 2;
    if (b >= n) break;
    long long k = i * (i + 1) / 2;
    if ((n - b) / k < i) break;
    if ((n - b) % k == 0) {
      ans[++tot].x = i;
      ans[tot].y = (n - b) / k;
      if (ans[tot].x == ans[tot].y) same++;
    }
    i++;
  }
  cout << 2 * tot - same << endl;
  for (int i = 1; i <= tot; i++) cout << ans[i].x << " " << ans[i].y << endl;
  for (int i = tot; i; i--)
    if (ans[i].x != ans[i].y) cout << ans[i].y << " " << ans[i].x << endl;
}

```