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() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long x;
  cin >> x;
  if (x == 1) {
    cout << 1 << endl;
    cout << 1 << " " << 1 << endl;
    return 0;
  }
  vector<pair<long long, long long> > res;
  long long d = 2 * cbrtl(x);
  for (long long i = 1; i <= d; i++) {
    long long n = i;
    long long m = (n - n * n * n - 6 * x) / (-3 * (n * n + n));
    long long mod = (n - n * n * n - 6 * x) % (-3 * (n * n + n));
    if (n > m) break;
    if (mod != 0) continue;
    res.push_back({n, m});
  }
  int n = res.size();
  for (int i = 0; i < n; i++) {
    if (res[i].first == res[i].second) continue;
    res.push_back({res[i].second, res[i].first});
  }
  sort(res.begin(), res.end());
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); i++) {
    cout << res[i].first << " " << res[i].second << endl;
  }
  return 0;
}