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 x;
int main() {
  scanf("%lld", &x);
  vector<pair<long long, long long> > res;
  long long to = sqrt(x);
  bool flag = false;
  for (long long i = 1; i <= to; i++) {
    long long s = x - (i - 1) * (i - 1 + 1ll) * (2 * (i - 1) + 1ll) / 6ll;
    s += i * (1 + (i - 1)) * (i - 1) / 2;
    long long temp = i * i - (1 + (i - 1)) * (i - 1) / 2;
    if (temp <= 0) break;
    if (s <= 0) break;
    if (s % temp == 0) {
      long long m = s / temp;
      if (m < i) break;
      if (i == m) flag = true;
      res.push_back(pair<long long, long long>(i, m));
    }
  }
  printf("%d\n", res.size() * 2 - flag);
  for (int i = 0; i < res.size(); i++)
    printf("%lld %lld\n", res[i].first, res[i].second);
  for (int i = (int)res.size() - 1; i >= 0; i--)
    if (res[i].first != res[i].second)
      printf("%lld %lld\n", res[i].second, res[i].first);
  return 0;
}