#include <bits/stdc++.h>
using namespace std;
int main(int argc, char const *argv[]) {
  long long x;
  scanf("%lld", &x);
  vector<pair<long long, long long> > ans;
  int f = 0;
  for (long long n = 1; n <= 2000010; n++) {
    long long k = x - n * n - n + ((n * (n + 1) * (n + 5)) / 6);
    long long t = n * (n + 1);
    t /= 2;
    if (k % t == 0) {
      if (k / t >= n) ans.push_back(pair<long long, long long>(n, k / t));
      if (n == k / t) f = 1;
    }
  }
  printf("%d\n", 2 * (int)ans.size() - f);
  for (int i = 0; i < ans.size(); i++) {
    printf("%lld %lld\n", ans[i].first, ans[i].second);
  }
  for (int i = ans.size() - f - 1; i >= 0; i--) {
    printf("%lld %lld\n", ans[i].second, ans[i].first);
  }
  return 0;
}

### Prompt

Generate 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 main(int argc, char const *argv[]) {
  long long x;
  scanf("%lld", &x);
  vector<pair<long long, long long> > ans;
  int f = 0;
  for (long long n = 1; n <= 2000010; n++) {
    long long k = x - n * n - n + ((n * (n + 1) * (n + 5)) / 6);
    long long t = n * (n + 1);
    t /= 2;
    if (k % t == 0) {
      if (k / t >= n) ans.push_back(pair<long long, long long>(n, k / t));
      if (n == k / t) f = 1;
    }
  }
  printf("%d\n", 2 * (int)ans.size() - f);
  for (int i = 0; i < ans.size(); i++) {
    printf("%lld %lld\n", ans[i].first, ans[i].second);
  }
  for (int i = ans.size() - f - 1; i >= 0; i--) {
    printf("%lld %lld\n", ans[i].second, ans[i].first);
  }
  return 0;
}

```