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.

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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>
const double PI = 2 * asin(1);
using namespace std;
long long n[2000000], m[2000000];
long long cal(long long x, long long n) {
  if ((6 * x) % (n * (n + 1)) == 0) return (6 * x / n / (n + 1) - 1 + n) / 3;
  return -1;
}
int main() {
  int index = 0;
  long long x;
  cin >> x;
  for (long long i = 1;; i++) {
    long long m1;
    if (i * (i + 1) * (2 * i + 1) / 6 > x) break;
    if ((m1 = cal(x, i)) != -1) {
      if (i > m1) break;
      if ((3 * m1 - i + 1) * i * (i + 1) / 6 == x) {
        n[index] = i;
        m[index] = m1;
        index++;
      }
    }
  }
  if (m[index - 1] == n[index - 1])
    cout << 2 * index - 1 << endl;
  else
    cout << 2 * index << endl;
  for (int i = 0; i < index; i++) printf("%lld %lld\n", n[i], m[i]);
  if (m[index - 1] == n[index - 1]) index--;
  for (int i = index - 1; i >= 0; i--) printf("%lld %lld\n", m[i], n[i]);
}