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;
const long long inf = (long long)1e9;
long long f(long long t, long long m) {
  return t * t * m - t * (t - 1) * (t + m) / 2;
}
long long work(double x, double t) {
  double tmp1 = (t * t * t - t * t) / 2 - x;
  double tmp2 = (t * t + t) / 2;
  double m = tmp1 / tmp2;
  return m + 0.5;
}
int main() {
  long long x;
  cin >> x;
  long long t = 0;
  vector<pair<long long, long long> > ans;
  for (long long i = 0; t <= x; i++) {
    t += i * i;
    long long tx = t - x;
    long long m = work(tx, i + 1);
    if (m >= i + 1 && f(i + 1, m) + tx == 0) {
      ans.push_back(pair<long long, long long>(i + 1, m));
      if (i + 1 != m) ans.push_back(pair<long long, long long>(m, i + 1));
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].first << " " << ans[i].second << endl;
  }
  return 0;
}