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;
long long s = 1;
long long n, m;
vector<pair<long long, long long> > V;
map<int, int> M;
long long S1(long long n) { return n * (n - 1) / 2; }
long long S2(long long n) {
  n--;
  return n * (n + 1) * (2 * n + 1) / 6;
}
int main() {
  long long x;
  scanf("%I64d", &x);
  long long n = 1, m;
  int cnt = 0;
  for (; n <= 2000000; n++) {
    if (S2(n - 1) > 1e18) break;
    if ((x + n * S1(n) - S2(n)) % (n * n - S1(n)) == 0)
      m = (x + n * S1(n) - S2(n)) / (n * n - S1(n));
    else
      continue;
    if (n > m) break;
    if (n == m) cnt++, M[n] = 1;
    V.push_back(make_pair(n, m));
  }
  printf("%I64d\n", 2 * (long long)V.size() - cnt);
  for (long long i = 0; i < V.size(); i++)
    printf("%I64d %I64d\n", V[i].first, V[i].second);
  for (long long i = V.size() - 1; i >= 0; i--) {
    if (M[V[i].first]) continue;
    printf("%I64d %I64d\n", V[i].second, V[i].first);
  }
  return 0;
}