#include <bits/stdc++.h>
using namespace std;
int k;
long long x, y, n, m;
pair<long long, long long> s[10000000 + 9];
int main() {
  scanf("%I64d", &x);
  for (long long i = 2; i < 10000000; ++i) {
    n = i;
    y = 6 * x + n * (n * n - 1);
    if (y % (3 * n * (n - 1)) == 0) {
      m = y / (3 * n * (n - 1));
      if (n == m) {
        s[k++] = make_pair(n - 1, m - 1);
      } else if (n < m) {
        s[k++] = make_pair(n - 1, m - 1);
        s[k++] = make_pair(m - 1, n - 1);
      }
    }
  }
  sort(s, s + k);
  printf("%d\n", k);
  for (int i = 0; i < k; ++i) printf("%I64d %I64d\n", s[i].first, s[i].second);
  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;
int k;
long long x, y, n, m;
pair<long long, long long> s[10000000 + 9];
int main() {
  scanf("%I64d", &x);
  for (long long i = 2; i < 10000000; ++i) {
    n = i;
    y = 6 * x + n * (n * n - 1);
    if (y % (3 * n * (n - 1)) == 0) {
      m = y / (3 * n * (n - 1));
      if (n == m) {
        s[k++] = make_pair(n - 1, m - 1);
      } else if (n < m) {
        s[k++] = make_pair(n - 1, m - 1);
        s[k++] = make_pair(m - 1, n - 1);
      }
    }
  }
  sort(s, s + k);
  printf("%d\n", k);
  for (int i = 0; i < k; ++i) printf("%I64d %I64d\n", s[i].first, s[i].second);
  return 0;
}

```