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 first;
int main() {
  scanf("%lld", &first);
  long long n = 1;
  vector<pair<long long, long long> > ans;
  while (n <= 1500000) {
    long long k = n - 1;
    long long top = first + ((k + 1LL) * k / 2LL) * n -
                    k * (k + 1LL) * (2LL * k + 1LL) / 6LL;
    long long bottom = (n * (k + 1LL) - (k + 1LL) * k / 2LL);
    if (bottom > 0 && top % bottom == 0) {
      long long m = top / bottom;
      if (m >= n) {
        ans.push_back(make_pair(n, m));
      }
    }
    n++;
  }
  set<pair<long long, long long> > rans;
  for (int(i) = (0); (i) < (int)(ans.size()); ++(i)) {
    rans.insert(make_pair(ans[i].first, ans[i].second));
    rans.insert(make_pair(ans[i].second, ans[i].first));
  }
  set<pair<long long, long long> >::iterator it = rans.begin();
  printf("%d\n", (int)rans.size());
  for (; it != rans.end(); it++) {
    printf("%lld %lld\n", it->first, it->second);
  }
  return 0;
}