#include <bits/stdc++.h>
using namespace std;
struct arr {
  long long x, y;
};
long long n, ans;
inline long long js(long long x) { return x * (x + 1) * (2 * x + 1) / 6; }
vector<arr> a;
signed main() {
  scanf("%lld", &n);
  long long i = 1;
  bool flag = 0;
  while (js(i) <= n && i <= 2000000) {
    long long sum = js(i), add = i * (i + 1) / 2;
    if ((n - sum) % add == 0) {
      a.push_back((arr){i, i + (n - sum) / add});
      ans++;
      if (n == sum) flag = 1;
    }
    i++;
  }
  ans <<= 1;
  if (flag) ans--;
  printf("%lld\n", ans);
  for (long long i = 0; i < a.size(); i++)
    printf("%lld %lld\n", a[i].x, a[i].y);
  for (long long i = a.size() - 1; i >= 0; i--)
    if (a[i].x != a[i].y) printf("%lld %lld\n", a[i].y, a[i].x);
  return 0;
}

### Prompt

Please create a solution in cpp 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;
struct arr {
  long long x, y;
};
long long n, ans;
inline long long js(long long x) { return x * (x + 1) * (2 * x + 1) / 6; }
vector<arr> a;
signed main() {
  scanf("%lld", &n);
  long long i = 1;
  bool flag = 0;
  while (js(i) <= n && i <= 2000000) {
    long long sum = js(i), add = i * (i + 1) / 2;
    if ((n - sum) % add == 0) {
      a.push_back((arr){i, i + (n - sum) / add});
      ans++;
      if (n == sum) flag = 1;
    }
    i++;
  }
  ans <<= 1;
  if (flag) ans--;
  printf("%lld\n", ans);
  for (long long i = 0; i < a.size(); i++)
    printf("%lld %lld\n", a[i].x, a[i].y);
  for (long long i = a.size() - 1; i >= 0; i--)
    if (a[i].x != a[i].y) printf("%lld %lld\n", a[i].y, a[i].x);
  return 0;
}

```