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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > soln;
void solve(long long x) {
  if (x == 1)
    printf("1\n1 1\n");
  else {
    long long a, b, i;
    for (i = 1;; ++i) {
      a = x + i * i * (i - 1) / 2 - i * (i - 1) * (2 * i - 1) / 6,
      b = i * i - i * (i - 1) / 2;
      if (a / b < i) break;
      if (a % b == 0) soln.push_back(pair<long long, long long>(i, a / b));
    }
    printf("%d\n", 2 * soln.size() - (soln[soln.size() - 1].first ==
                                      soln[soln.size() - 1].second));
    for (i = 0; i < soln.size(); ++i)
      printf("%I64d %I64d\n", soln[i].first, soln[i].second);
    for (i = (soln[soln.size() - 1].first == soln[soln.size() - 1].second
                  ? soln.size() - 2
                  : soln.size() - 1);
         i >= 0; --i)
      printf("%I64d %I64d\n", soln[i].second, soln[i].first);
  }
}
int main() {
  long long x;
  scanf("%I64d", &x);
  solve(x);
  return 0;
}