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 sol(long long n, long long t) {
  long long ret = n * (n + 1) * (2 * n + 3 * t + 1);
  ret /= 6;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long x;
  cin >> x;
  long long ans = 0;
  vector<pair<long long, long long> > aa;
  for (long long n = 1;; n++) {
    long long z = sol(n, 0);
    if (z > x) {
      break;
    }
    long long r = 6 * x - 2 * n * n * n - 3 * n * n - n;
    long long d = 3 * n * n + 3 * n;
    if (r % d == 0) {
      ans++;
      aa.push_back(make_pair(n, n + r / d));
    }
  }
  long long xx = aa.size();
  for (int i = 0; i < xx; i++) {
    if (aa[i].first != aa[i].second)
      aa.push_back(make_pair(aa[i].second, aa[i].first));
  }
  cout << aa.size() << "\n";
  sort(aa.begin(), aa.end());
  for (int i = 0; i < aa.size(); i++) {
    cout << aa[i].first << " " << aa[i].second << "\n";
  }
  return 0;
}