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;
inline void fastIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
vector<pair<long long, long long> > v;
const int Maxn = 200005;
void Print(vector<pair<long long, long long> > &P) {
  cout << P.size() << endl;
  for (long long i = 0; i < P.size(); i++)
    cout << P[i].first << " " << P[i].second << endl;
}
int main() {
  fastIO();
  long long X;
  cin >> X;
  X *= 3;
  for (long long i = 1; i * i * i <= X; i++) {
    long long Z = (i * (i + 1) / 2);
    if (X % Z == 0) {
      long long Temp = X / Z;
      if ((Temp + i - 1) % 3 == 0) {
        long long Tm = (Temp + i - 1) / 3;
        if (Tm >= i) {
          v.push_back(pair<long long, long long>(i, Tm));
          if (i != Tm) v.push_back(pair<long long, long long>(Tm, i));
        }
      }
    }
  }
  sort(v.begin(), v.end());
  Print(v);
  return 0;
}