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.

<image>

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;
const int mod = 1e9 + 7;
const double pi = 3.1415926536;
const double eps = 1e-9;
const int dX[] = {0, 0, 1, -1, 1, -1, 1, -1};
const int dY[] = {1, -1, 0, 0, 1, -1, -1, 1};
inline long long modPow(long long b, long long e) {
  long long s = 1;
  while (e) {
    if (e & 1) s = (s * b) % mod;
    b = (b * b) % mod;
    e >>= 1;
  }
  return s;
}
inline void primeGen(vector<bool> &p) {
  int i, j;
  for (i = (2); (((1) > 0) - ((1) < 0)) * i <
                (((1) > 0) - ((1) < 0)) * (((long long int)(p).size()));
       i += (1)) {
    if (p[i])
      for (j = (i + i); (((i) > 0) - ((i) < 0)) * j <
                        (((i) > 0) - ((i) < 0)) * (((long long int)(p).size()));
           j += (i))
        p[j] = 0;
  }
}
inline bool isInt(double x) { return x == floor(x); }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  long long i = 1, n, m, t;
  set<pair<long long, long long> > ans;
  cin >> n;
  for (i = (1);
       (((1) > 0) - ((1) < 0)) * i < (((1) > 0) - ((1) < 0)) * (1500100);
       i += (1)) {
    m = (n + ((i) * (i) * (i - 1)) / 2 - (i * (i - 1) * (2 * i - 1)) / 6);
    t = ((i * (i + 1)) / 2);
    if (!(m % t)) {
      m /= t;
      if (m <= 0 || m < i) continue;
      ans.insert(make_pair(i, m));
      if (i != m) ans.insert(make_pair(m, i));
    }
  }
  cout << ((long long int)(ans).size()) << "\n";
  for (typeof((ans).begin()) it = (ans).begin(); it != (ans).end(); it++)
    cout << it->first << " " << it->second << "\n";
  return 0;
}