#include <bits/stdc++.h>
using namespace std;
long long x, ans;
struct pp {
  long long n, m;
} p[8000005];
bool cmp(const pp &i, const pp &j) { return i.n < j.n; }
int main() {
  scanf("%I64d", &x);
  x *= 6;
  for (long long i = 1; i <= 4000000; i++) {
    if (x % (i + 1ll) == 0) {
      long long td = x / (i + 1ll);
      td = td + i * i - i;
      if (td % (i * 3ll) == 0) {
        long long m = td / (i * 3ll);
        if (i <= m) {
          p[++ans].n = i;
          p[ans].m = m;
          if (i != m) {
            p[++ans].n = m;
            p[ans].m = i;
          }
        }
      }
    }
  }
  sort(p + 1, p + ans + 1, cmp);
  printf("%I64d\n", ans);
  for (int i = 1; i <= ans; i++) printf("%I64d %I64d\n", p[i].n, p[i].m);
}

### 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;
long long x, ans;
struct pp {
  long long n, m;
} p[8000005];
bool cmp(const pp &i, const pp &j) { return i.n < j.n; }
int main() {
  scanf("%I64d", &x);
  x *= 6;
  for (long long i = 1; i <= 4000000; i++) {
    if (x % (i + 1ll) == 0) {
      long long td = x / (i + 1ll);
      td = td + i * i - i;
      if (td % (i * 3ll) == 0) {
        long long m = td / (i * 3ll);
        if (i <= m) {
          p[++ans].n = i;
          p[ans].m = m;
          if (i != m) {
            p[++ans].n = m;
            p[ans].m = i;
          }
        }
      }
    }
  }
  sort(p + 1, p + ans + 1, cmp);
  printf("%I64d\n", ans);
  for (int i = 1; i <= ans; i++) printf("%I64d %I64d\n", p[i].n, p[i].m);
}

```