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;
long long get(long long n, long long m) {
  long long ans = 0;
  for (long long k = 0; k < min(n, m); k++) ans += (n - k) * (m - k);
  return ans;
}
long long get2(long long n, long long m) {
  if (n > m) {
    long long t = n;
    n = m;
    m = t;
  }
  return m * (n * n + n) / 2 - n * n * (n + 1) / 2 +
         n * (n + 1) * (2 * n + 1) / 6;
}
long long get3(long long n, long long x) {
  long long left = x + n * n * (n + 1) / 2 - n * (n + 1) * (2 * n + 1) / 6;
  left *= 2;
  left /= n * n + n;
  if (left >= n && get2(n, left) == x)
    return left;
  else
    return -1;
}
int main() {
  ios::sync_with_stdio(false);
  long long x;
  cin >> x;
  vector<pair<long long, long long> > ans;
  for (long long n = 1; n <= 1500000; n++) {
    long long m = get3(n, x);
    if (m != -1) {
      ans.push_back(make_pair(n, m));
      if (n != m) ans.push_back(make_pair(m, n));
    }
  }
  sort(ans.begin(), ans.end());
  int s = ans.size();
  cout << s << endl;
  for (int i = 0; i < s; i++)
    cout << ans[i].first << " " << ans[i].second << endl;
  return 0;
}