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 int func1(long long int i) {
  long long int m = i;
  return i * (i + 1) * (2 * i + 1) / 6 - (m + m + 2) * i * (i + 1) / 2 +
         (m + m + 1 + m * m) * i;
}
long long int func2(long long int x, long long int i) {
  return x * (i + i * i - i * (i + 1) / 2);
}
int main() {
  long long int inp, tem, ans = 0, tem2;
  cin >> inp;
  vector<pair<long long int, long long int> > v;
  bool flag = 0;
  for (int i = 0; i < 100000000; i++)
    if (ans > 0) ans += 1;
  ans = 0;
  for (long long int i = 1; func1(i) <= inp; i++) {
    tem = inp - func1(i);
    if (!tem) {
      flag = 1;
      v.push_back(make_pair(i, i));
    } else {
      tem2 = func2(1, i);
      if (tem % tem2 == 0) v.push_back(make_pair(i, i + tem / tem2));
    }
  }
  if (!flag)
    cout << 2 * v.size();
  else
    cout << 2 * v.size() - 1;
  cout << '\n';
  for (int i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << '\n';
  int i = v.size() - 1;
  if (flag) i--;
  for (; i >= 0; i--) cout << v[i].second << " " << v[i].first << '\n';
  return 0;
}