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;
long long x, p, q, w;
int main() {
  while (~scanf("%I64d", &x)) {
    vector<long long> a, b;
    long long tot = 0;
    for (long long i = 1; i * (i + 1) * (2 * i + 1) <= 6 * x; i++) {
      if (6 * x % i == 0) {
        p = 6 * x / i;
        if (p % (i + 1) == 0) {
          p = p / (i + 1);
          if (p - 2 * i - 1 >= 0) {
            p = p - (2 * i + 1);
            if (p % 3 == 0) {
              p = p / 3;
              a.push_back(i);
              b.push_back(i + p);
            }
          }
        }
      }
    }
    tot = a.size();
    if (a[tot - 1] == b[tot - 1])
      printf("%I64d\n", tot * 2 - 1);
    else
      printf("%I64d\n", 2 * tot);
    for (long long i = 0; i < tot; i++) printf("%I64d %I64d\n", a[i], b[i]);
    for (long long i = tot - 1; i >= 0; i--) {
      if (a[i] == b[i]) continue;
      printf("%I64d %I64d\n", b[i], a[i]);
    }
  }
  return 0;
}