#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > S;
int main() {
  long long x, i, Ans = 0;
  scanf("%lld", &x);
  for (i = 1;; ++i) {
    long long n = i, m = (6LL * x - n + n * n * n) / (3 * n * n + 3 * n);
    if (n > m) break;
    if ((6LL * x - n + n * n * n) % (3 * n * n + 3 * n) == 0) {
      S.push_back(make_pair(n, m));
    }
  }
  if (S.back().first == S.back().second) {
    Ans = -1;
  }
  Ans += (int)S.size() << 1;
  printf("%lld\n", Ans);
  for (i = 0; i < (int)S.size(); ++i) {
    printf("%lld %lld\n", S[i].first, S[i].second);
  }
  if (S.back().first == S.back().second) {
    S.pop_back();
  }
  while (!S.empty()) {
    printf("%lld %lld\n", S.back().second, S.back().first);
    S.pop_back();
  }
  return 0;
}

### 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>
using namespace std;
vector<pair<long long, long long> > S;
int main() {
  long long x, i, Ans = 0;
  scanf("%lld", &x);
  for (i = 1;; ++i) {
    long long n = i, m = (6LL * x - n + n * n * n) / (3 * n * n + 3 * n);
    if (n > m) break;
    if ((6LL * x - n + n * n * n) % (3 * n * n + 3 * n) == 0) {
      S.push_back(make_pair(n, m));
    }
  }
  if (S.back().first == S.back().second) {
    Ans = -1;
  }
  Ans += (int)S.size() << 1;
  printf("%lld\n", Ans);
  for (i = 0; i < (int)S.size(); ++i) {
    printf("%lld %lld\n", S[i].first, S[i].second);
  }
  if (S.back().first == S.back().second) {
    S.pop_back();
  }
  while (!S.empty()) {
    printf("%lld %lld\n", S.back().second, S.back().first);
    S.pop_back();
  }
  return 0;
}

```