#include <bits/stdc++.h>
using namespace std;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  n *= 6;
  vector<pair<long long, long long> > res;
  for (long long b = 1;; b++) {
    long long t = b * (b + 1) * (2 * b + 1);
    if (t > n) break;
    if (n % (b * (b + 1))) continue;
    long long x = n / b / (b + 1);
    long long a = x - 1 - b * 2;
    if (a % 3 == 0) {
      a /= 3;
      res.push_back(make_pair(b, b + a));
      if (a) res.push_back(make_pair(a + b, b));
    }
  }
  sort((res).begin(), (res).end());
  cout << ((int)(res).size()) << endl;
  for (int i = 0; i < (((int)(res).size())); ++i)
    cout << res[i].first << " " << res[i].second << "\n";
  return 0;
}

### Prompt

Please formulate 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;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  n *= 6;
  vector<pair<long long, long long> > res;
  for (long long b = 1;; b++) {
    long long t = b * (b + 1) * (2 * b + 1);
    if (t > n) break;
    if (n % (b * (b + 1))) continue;
    long long x = n / b / (b + 1);
    long long a = x - 1 - b * 2;
    if (a % 3 == 0) {
      a /= 3;
      res.push_back(make_pair(b, b + a));
      if (a) res.push_back(make_pair(a + b, b));
    }
  }
  sort((res).begin(), (res).end());
  cout << ((int)(res).size()) << endl;
  for (int i = 0; i < (((int)(res).size())); ++i)
    cout << res[i].first << " " << res[i].second << "\n";
  return 0;
}

```