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;
int main() {
  long long unsigned x;
  scanf("%I64u", &x);
  long long unsigned y = 6 * x, z;
  vector<pair<long long unsigned, long long unsigned> > r;
  for (long long unsigned i = 1; i < 1500000; ++i) {
    if ((2 * i + 1) * i * (i + 1) / 6 > x) break;
    if (y % (i * (i + 1)) != 0) continue;
    z = y / i / (i + 1) + i - 1;
    if (z % 3 != 0) continue;
    r.push_back(make_pair(i, z / 3));
  }
  bool c = r.back().first == r.back().second;
  if (c) {
    printf("%d\n", r.size() * 2 - 1);
  } else {
    printf("%d\n", r.size() * 2);
  }
  for (int i = 0; i < r.size(); ++i)
    printf("%I64u %I64u\n", r[i].first, r[i].second);
  for (int i = (c ? ((int)r.size() - 2) : r.size() - 1); i >= 0; --i) {
    printf("%I64u %I64u\n", r[i].second, r[i].first);
  }
  return 0;
}