#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > ans, ans2;
int main() {
  long long n;
  scanf("%lld", &n);
  long long sum1 = 0;
  long long sum2 = 0;
  for (long long i = 1; i <= 1500000; i++) {
    sum1 += i;
    sum2 += i * (i - 1);
    if ((n - sum2) % sum1 == 0) {
      long long y = (n - sum2) / sum1 + (i - 1);
      if (y < i) break;
      ans.push_back({i, y});
      if (y == i) break;
      ans2.push_back({y, i});
    }
  }
  printf("%d\n", ans.size() + ans2.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%lld %lld\n", ans[i].first, ans[i].second);
  }
  for (int i = ans2.size() - 1; i >= 0; i--) {
    printf("%lld %lld\n", ans2[i].first, ans2[i].second);
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
vector<pair<long long, long long> > ans, ans2;
int main() {
  long long n;
  scanf("%lld", &n);
  long long sum1 = 0;
  long long sum2 = 0;
  for (long long i = 1; i <= 1500000; i++) {
    sum1 += i;
    sum2 += i * (i - 1);
    if ((n - sum2) % sum1 == 0) {
      long long y = (n - sum2) / sum1 + (i - 1);
      if (y < i) break;
      ans.push_back({i, y});
      if (y == i) break;
      ans2.push_back({y, i});
    }
  }
  printf("%d\n", ans.size() + ans2.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%lld %lld\n", ans[i].first, ans[i].second);
  }
  for (int i = ans2.size() - 1; i >= 0; i--) {
    printf("%lld %lld\n", ans2[i].first, ans2[i].second);
  }
  return 0;
}

```