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;
set<pair<long long, long long> > m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, a, b, c, d, e, l = 0;
  cin >> n;
  for (long long i = 1; i <= 10000000; i++) {
    a = (i * (i + 1)) / 2;
    b = n;
    c = b / a;
    d = (i * (i - 1) * (i + 1)) / 6;
    if ((n % a + d % a) % a == 0) {
      long long p;
      if (n % a == 0) {
        p = n / a;
        p += d / a;
      }
      if (n % a != 0) {
        p = n / a;
        p += d / a + 1;
      }
      if (i > p) break;
      m.insert(make_pair(i, p));
      m.insert(make_pair(p, i));
    }
    if (b > 1000000000000000000) break;
  }
  cout << m.size() << endl;
  for (set<pair<long long, long long> >::iterator i = m.begin(); i != m.end();
       i++) {
    cout << (*i).first << " " << (*i).second << endl;
  }
}