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, tmp;
vector<pair<long long, long long> > ans;
long double q;
int main() {
  cin >> x;
  ans.push_back(make_pair(1, x));
  if (x != 1) ans.push_back(make_pair(x, 1));
  for (int n = 2; n < 1817121; ++n) {
    q = n + 1;
    q *= n;
    q *= n - 1;
    q /= 6;
    if (q > x) break;
    tmp = x;
    tmp -= (long long)q;
    tmp -= (long long)q;
    if (tmp % (n * 1ll * (n + 1) / 2) == 0) {
      long long diff = tmp / (n * 1ll * (n + 1) / 2);
      long long m = n + diff - 1;
      if (m >= n) {
        ans.push_back(make_pair(n, m));
        if (m != n) ans.push_back(make_pair(m, n));
      }
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); ++i)
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  return 0;
}