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 int max_b;
long long int get_size(long long int x, long long int a) {
  long long int orig = a * (a + 1) * (2 * a + 1) / 6;
  if (orig > x) return -1;
  long long int k = 2 * (x - orig);
  long long int denom = a * (a + 1);
  max_b = a + k / denom;
  if (k % denom == 0) {
    return max_b;
  } else
    return -1;
}
int main() {
  long long int x;
  cin >> x;
  if (x == 1) {
    cout << "1" << endl << "1 1" << endl;
    return 0;
  }
  long long int a = 1;
  max_b = x;
  vector<pair<long long int, long long int>> ans;
  long long int sum = 0;
  while (a < max_b) {
    long long int b = get_size(x, a);
    if (b != -1) {
      ans.push_back(make_pair(a, b));
      max_b = b;
      sum++;
      if (a != b) sum++;
    }
    a++;
  }
  cout << sum << endl;
  if (sum != 0) {
    for (auto &x : ans) {
      cout << x.first << " " << x.second << endl;
    }
    for (int i = (int)ans.size() - 1; i >= 0; i--) {
      if (ans[i].first != ans[i].second) {
        cout << ans[i].second << " " << ans[i].first << endl;
      }
    }
  }
}