#include <bits/stdc++.h>
template <class T>
inline T gcd(T a, T b) {
  while (b) b ^= a ^= b ^= a %= b;
  return a;
}
template <class T>
inline T pow(T a, T b, T c) {
  T re = 1;
  while (b) {
    if (b & 1) re = (re * a) % c;
    a = (a * a) % c;
    b >>= 1;
  }
  return re;
}
template <class T>
inline void read(T& x) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (x = 0; c > 32; c = getchar()) x = x * 10 + c - '0';
  if (fu) x = -x;
};
template <class T>
inline void read(T& x, T& y) {
  read(x);
  read(y);
}
template <class T>
inline void read(T& x, T& y, T& z) {
  read(x);
  read(y);
  read(z);
}
using namespace std;
vector<pair<long long, long long> > v;
int main() {
  long long x, root, m, a, b;
  map<long long, bool> A;
  read(x);
  long long r = 1, x1 = x;
  while (r * r * r <= x) r++;
  for (long long i = 1; i < 2 * r; ++i) {
    if (6 * x1 % i == 0) {
      x = 6 * x1 / i;
      a = (x - 1 + i * i);
      b = 3 * i + 3;
      if (a % b == 0) {
        m = a / b;
        if (A[m] or m < i) continue;
        A[m] = true, A[i] = true;
        if (i == m)
          v.push_back(make_pair(i, m));
        else {
          v.push_back(make_pair(i, m));
          v.push_back(make_pair(m, i));
        }
      }
    }
  }
  sort(v.begin(), v.end());
  cout << v.size() << endl;
  for (long long i = 0; i < v.size(); i += 1)
    cout << v[i].first << ' ' << v[i].second << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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>
template <class T>
inline T gcd(T a, T b) {
  while (b) b ^= a ^= b ^= a %= b;
  return a;
}
template <class T>
inline T pow(T a, T b, T c) {
  T re = 1;
  while (b) {
    if (b & 1) re = (re * a) % c;
    a = (a * a) % c;
    b >>= 1;
  }
  return re;
}
template <class T>
inline void read(T& x) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (x = 0; c > 32; c = getchar()) x = x * 10 + c - '0';
  if (fu) x = -x;
};
template <class T>
inline void read(T& x, T& y) {
  read(x);
  read(y);
}
template <class T>
inline void read(T& x, T& y, T& z) {
  read(x);
  read(y);
  read(z);
}
using namespace std;
vector<pair<long long, long long> > v;
int main() {
  long long x, root, m, a, b;
  map<long long, bool> A;
  read(x);
  long long r = 1, x1 = x;
  while (r * r * r <= x) r++;
  for (long long i = 1; i < 2 * r; ++i) {
    if (6 * x1 % i == 0) {
      x = 6 * x1 / i;
      a = (x - 1 + i * i);
      b = 3 * i + 3;
      if (a % b == 0) {
        m = a / b;
        if (A[m] or m < i) continue;
        A[m] = true, A[i] = true;
        if (i == m)
          v.push_back(make_pair(i, m));
        else {
          v.push_back(make_pair(i, m));
          v.push_back(make_pair(m, i));
        }
      }
    }
  }
  sort(v.begin(), v.end());
  cout << v.size() << endl;
  for (long long i = 0; i < v.size(); i += 1)
    cout << v[i].first << ' ' << v[i].second << endl;
  return 0;
}

```