#include <bits/stdc++.h>
using namespace std;
struct per {
  long long n, m;
};
long long x, K;
per ans[1200100];
int main() {
  cin >> x;
  long long n = 1, s = 1, aux, m, val;
  while (s <= x) {
    aux = x - s;
    val = n * (n + 1) / 2;
    if (aux % val == 0) {
      m = aux / val;
      K++;
      ans[K].n = n;
      ans[K].m = n + m;
    }
    n++;
    s += n * n;
    if (s < 0) break;
  }
  aux = K;
  long long i;
  for (i = aux; i >= 1; i--) {
    if (ans[i].n != ans[i].m) {
      K++;
      ans[K].n = ans[i].m;
      ans[K].m = ans[i].n;
    }
  }
  cout << K << "\n";
  for (i = 1; i <= K; i++) cout << ans[i].n << " " << ans[i].m << "\n";
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 per {
  long long n, m;
};
long long x, K;
per ans[1200100];
int main() {
  cin >> x;
  long long n = 1, s = 1, aux, m, val;
  while (s <= x) {
    aux = x - s;
    val = n * (n + 1) / 2;
    if (aux % val == 0) {
      m = aux / val;
      K++;
      ans[K].n = n;
      ans[K].m = n + m;
    }
    n++;
    s += n * n;
    if (s < 0) break;
  }
  aux = K;
  long long i;
  for (i = aux; i >= 1; i--) {
    if (ans[i].n != ans[i].m) {
      K++;
      ans[K].n = ans[i].m;
      ans[K].m = ans[i].n;
    }
  }
  cout << K << "\n";
  for (i = 1; i <= K; i++) cout << ans[i].n << " " << ans[i].m << "\n";
  return 0;
}

```