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;
int main() {
  vector<pair<long long, long long> > ans;
  long long n, i, t, m, val1, val2, cnt(0);
  scanf("%I64d", &n);
  t = n * 6;
  for (i = 1; i * i * i <= t; i++) {
    if (t % i == 0) {
      val1 = t + i * i * i - i;
      val2 = i * 3 + i * i * 3;
      if (val1 % val2 == 0) {
        m = val1 / val2;
        if (i <= m) {
          cnt++;
          ans.push_back(make_pair(i, m));
          if (i != m) {
            cnt++;
          }
        }
      }
    }
  }
  printf("%I64d\n", cnt);
  for (i = 0; i < (int(ans.size())); i++) {
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  }
  for (i = (int(ans.size())) - 1; i >= 0; i--) {
    if (ans[i].first != ans[i].second) {
      printf("%I64d %I64d\n", ans[i].second, ans[i].first);
    }
  }
  return 0;
}