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>
using namespace std;
vector<pair<unsigned long long, unsigned long long> > ans;
int main() {
  unsigned long long x;
  scanf("%I64d", &x);
  x = 6 * x;
  unsigned long long lim = (unsigned long long)cbrt(x / 2.0);
  unsigned long long special = -1;
  for (unsigned long long i = 1; i <= lim; i++)
    if (x % i == 0) {
      unsigned long long temp = x / i;
      temp += i * i - 1;
      if (temp % (3 * i + 3) == 0) {
        temp /= (3 * i + 3);
        if (i == temp) {
          special = i;
        } else if (i < temp)
          ans.push_back(pair<unsigned long long, unsigned long long>(i, temp));
      }
    }
  int nr = 2 * ans.size();
  if (special != -1) nr++;
  printf("%d\n", nr);
  for (int i = 0; i < ans.size(); i++)
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  if (special != -1) printf("%I64d %I64d\n", special, special);
  for (int i = ans.size() - 1; i >= 0; i--)
    printf("%I64d %I64d\n", ans[i].second, ans[i].first);
  return 0;
}