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;
vector<long long> a(5123456);
int main() {
  std::ios::sync_with_stdio(false);
  for (int i = 0; i < 5123456; i++) a[i] = 0;
  a[1] = 1;
  for (int i = 2; i < 5123456; i++) {
    a[i] = a[i - 1] + ((long long)i * (i + 1)) / 2;
  }
  vector<pair<long long, long long> > fa;
  long long k;
  cin >> k;
  long long ans = k + 1;
  for (int i = 1; i < 5123456; i++) {
    long long n = ((long long)i * (i + 1)) / 2;
    long long other = a[i - 1];
    other += k;
    if (other % n == 0) {
      ans = other / n;
      if (ans < i) break;
      fa.push_back(make_pair(i, other / n));
    }
  }
  int nn = fa.size();
  for (int i = nn - 1; i >= 0; i--) {
    if (fa[i].first != fa[i].second)
      fa.push_back(make_pair(fa[i].second, fa[i].first));
  }
  cout << fa.size() << endl;
  for (int i = 0; i < fa.size(); i++)
    cout << fa[i].first << " " << fa[i].second << endl;
  return 0;
}