#include <bits/stdc++.h>
long long int x[1000000], y[1000000];
int main() {
  long long int n, i = 1, a = 0, b = 0;
  int ans = 0;
  scanf("%I64d", &n);
  while (true) {
    b += a;
    a += i;
    if ((n + b) / a < i) break;
    if ((n + b) % a == 0) {
      x[ans] = i;
      y[ans] = (n + b) / a;
      ans++;
    }
    i++;
  }
  if (x[ans - 1] != y[ans - 1])
    printf("%d\n", 2 * ans);
  else
    printf("%d\n", 2 * ans - 1);
  for (int j = 0; j < ans; j++) printf("%I64d %I64d\n", x[j], y[j]);
  if (x[ans - 1] != y[ans - 1]) printf("%I64d %I64d\n", y[ans - 1], x[ans - 1]);
  for (int j = ans - 2; j >= 0; j--) printf("%I64d %I64d\n", y[j], x[j]);
}

### Prompt

Develop a solution in CPP 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>
long long int x[1000000], y[1000000];
int main() {
  long long int n, i = 1, a = 0, b = 0;
  int ans = 0;
  scanf("%I64d", &n);
  while (true) {
    b += a;
    a += i;
    if ((n + b) / a < i) break;
    if ((n + b) % a == 0) {
      x[ans] = i;
      y[ans] = (n + b) / a;
      ans++;
    }
    i++;
  }
  if (x[ans - 1] != y[ans - 1])
    printf("%d\n", 2 * ans);
  else
    printf("%d\n", 2 * ans - 1);
  for (int j = 0; j < ans; j++) printf("%I64d %I64d\n", x[j], y[j]);
  if (x[ans - 1] != y[ans - 1]) printf("%I64d %I64d\n", y[ans - 1], x[ans - 1]);
  for (int j = ans - 2; j >= 0; j--) printf("%I64d %I64d\n", y[j], x[j]);
}

```