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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<long long> x, y;
inline long long Compare(const long long &a, const long long &b,
                         const long long &n) {
  return 6 * n - 3 * a * a * b + a * a * a - 3 * a * b - a;
}
long long Gao(const long long &n, const long long &a) {
  if (n < a) {
    return -1;
  }
  long long b = (6 * n + a * a * a - a) / (3 * a * a + 3 * a);
  long long res = Compare(a, b, n);
  if (0 == res) {
    return b;
  } else {
    return -1;
  }
}
int main() {
  long long n;
  ios::sync_with_stdio(0);
  while (cin >> n) {
    x.clear();
    y.clear();
    x.push_back(1);
    y.push_back(n);
    for (long long a = 2; a <= 3000000; ++a) {
      long long b = Gao(n, a);
      if (b > 0) {
        if (a > b) {
          break;
        }
        x.push_back(a);
        y.push_back(b);
      }
    }
    if (*x.rbegin() != *y.rbegin()) {
      cout << (x.size() * 2) << endl;
      for (int i = 0; i < x.size(); ++i) {
        cout << x[i] << ' ' << y[i] << endl;
      }
    } else {
      cout << (x.size() * 2 - 1) << endl;
      for (int i = 0; i < x.size() - 1; ++i) {
        cout << x[i] << ' ' << y[i] << endl;
      }
    }
    for (int i = x.size() - 1; i >= 0; --i) {
      cout << y[i] << ' ' << x[i] << endl;
    }
  }
  return 0;
}