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<long long, long long> > ans;
vector<pair<long long, long long> > ans2;
int main() {
  unsigned long long n;
  cin >> n;
  unsigned long long last = 1;
  unsigned long long mi = 0;
  while (last <= (int)1e7 && last <= n) {
    mi += (last) * (last - 1) / 2;
    unsigned long long di = (last * (last + 1)) / 2;
    if ((n + mi) % di == 0) {
      unsigned long long res = (n + mi) / di;
      if (res == last) {
        ans.push_back(make_pair(last, last));
        break;
      } else if (res < last)
        break;
      else {
        ans.push_back(make_pair(last, res));
        ans2.push_back(make_pair(res, last));
      }
    }
    last++;
  }
  printf("%d\n", ans.size() + ans2.size());
  for (int i = 0; i < ans.size(); ++i)
    printf("%lld %lld\n", ans[i].first, ans[i].second);
  reverse(ans2.begin(), ans2.end());
  for (int i = 0; i < ans2.size(); ++i)
    printf("%lld %lld\n", ans2[i].first, ans2[i].second);
  return 0;
}