#include <bits/stdc++.h>
using namespace std;
const int inf_int = 2e9;
const long long inf_ll = 2e18;
int Gcd(int p, int q) { return q == 0 ? p : Gcd(q, p % q); }
int Pow(int p, int q) {
  int ans = 1;
  while (q) {
    if (q & 1) ans = ans * p;
    p = p * p;
    q >>= 1;
  }
  return ans;
}
inline int read() {
  int ra, fh;
  char rx;
  rx = getchar(), ra = 0, fh = 1;
  while ((rx < '0' || rx > '9') && rx != '-') rx = getchar();
  if (rx == '-') fh = -1, rx = getchar();
  while (rx >= '0' && rx <= '9') ra *= 10, ra += rx - 48, rx = getchar();
  return ra * fh;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long x;
  cin >> x;
  x *= 6;
  vector<pair<long long, long long>> ans;
  for (long long n = 1; n <= 2000000; ++n) {
    long long t = n * (n + 1);
    if (x % t) continue;
    long long y = x / t - n - n - 1;
    if (y % 3) continue;
    y /= 3;
    if (y >= 0) {
      ans.push_back({n, n + y});
      ans.push_back({n + y, n});
    }
  }
  sort(ans.begin(), ans.end());
  ans.resize(std::unique(ans.begin(), ans.end()) - ans.begin());
  cout << ans.size() << endl;
  for (auto p : ans) {
    cout << p.first << ' ' << p.second << '\n';
  }
  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;
const int inf_int = 2e9;
const long long inf_ll = 2e18;
int Gcd(int p, int q) { return q == 0 ? p : Gcd(q, p % q); }
int Pow(int p, int q) {
  int ans = 1;
  while (q) {
    if (q & 1) ans = ans * p;
    p = p * p;
    q >>= 1;
  }
  return ans;
}
inline int read() {
  int ra, fh;
  char rx;
  rx = getchar(), ra = 0, fh = 1;
  while ((rx < '0' || rx > '9') && rx != '-') rx = getchar();
  if (rx == '-') fh = -1, rx = getchar();
  while (rx >= '0' && rx <= '9') ra *= 10, ra += rx - 48, rx = getchar();
  return ra * fh;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long x;
  cin >> x;
  x *= 6;
  vector<pair<long long, long long>> ans;
  for (long long n = 1; n <= 2000000; ++n) {
    long long t = n * (n + 1);
    if (x % t) continue;
    long long y = x / t - n - n - 1;
    if (y % 3) continue;
    y /= 3;
    if (y >= 0) {
      ans.push_back({n, n + y});
      ans.push_back({n + y, n});
    }
  }
  sort(ans.begin(), ans.end());
  ans.resize(std::unique(ans.begin(), ans.end()) - ans.begin());
  cout << ans.size() << endl;
  for (auto p : ans) {
    cout << p.first << ' ' << p.second << '\n';
  }
  return 0;
}

```