#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long>> ans;
int main() {
  long long x;
  cin >> x;
  for (long long i = 1; i <= cbrt(6 * x); i++) {
    long long c1 = i * (i + 1) * (2 * i + 1) / 6;
    if (c1 > x) break;
    long long c2 = i * (i + 1) / 2;
    if ((x - c1) % c2 == 0) ans.push_back({i, (x - c1) / c2 + i});
  }
  long long n = 2 * ans.size();
  if (ans[ans.size() - 1].first == ans[ans.size() - 1].second) n--;
  printf("%I64d\n", n);
  for (auto& p : ans) printf("%I64d %I64d\n", p.first, p.second);
  reverse(ans.begin(), ans.end());
  for (auto& p : ans)
    if (p.first != p.second) printf("%I64d %I64d\n", p.second, p.first);
  return 0;
}

### Prompt

Your task is to create 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;
vector<pair<long long, long long>> ans;
int main() {
  long long x;
  cin >> x;
  for (long long i = 1; i <= cbrt(6 * x); i++) {
    long long c1 = i * (i + 1) * (2 * i + 1) / 6;
    if (c1 > x) break;
    long long c2 = i * (i + 1) / 2;
    if ((x - c1) % c2 == 0) ans.push_back({i, (x - c1) / c2 + i});
  }
  long long n = 2 * ans.size();
  if (ans[ans.size() - 1].first == ans[ans.size() - 1].second) n--;
  printf("%I64d\n", n);
  for (auto& p : ans) printf("%I64d %I64d\n", p.first, p.second);
  reverse(ans.begin(), ans.end());
  for (auto& p : ans)
    if (p.first != p.second) printf("%I64d %I64d\n", p.second, p.first);
  return 0;
}

```