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;
const long long int MOD = 1e9 + 7;
const long double PI = 2 * acos(0.0);
const vector<long long int> dx = {1, -1, 0, 0};
const vector<long long int> dy = {0, 0, 1, -1};
template <typename T = long long int>
vector<T> ga(long long int n, bool oneIndexed = false) {
  vector<T> a = vector<T>(n + oneIndexed);
  for (long long int i = 0; i < n; i++) {
    T p;
    cin >> p;
    a[i + oneIndexed] = p;
  }
  return move(a);
}
template <typename T, typename A>
void pa(vector<T, A> const &a, long long int begin = 0,
        long long int end = -1) {
  if (end == -1) end = (long long int)a.size() - 1;
  for (long long int i = begin; i <= end; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
void solve() {
  long long int x;
  cin >> x;
  x *= 6;
  set<pair<long long int, long long int>> ans;
  for (long long int n = 1; n <= 2e6; n++) {
    long long int left = n * (n + 1);
    if (x % (left) == 0) {
      long long int right = x / left;
      right += n - 1;
      if (right % 3 == 0 && (right / 3) >= n) {
        ans.insert({n, right / 3});
        ans.insert({right / 3, n});
      }
    }
  }
  cout << (long long int)ans.size() << "\n";
  for (auto x : ans) cout << x.first << " " << x.second << "\n";
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solve();
  return 0;
}