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;
vector<pair<long long, long long> > fuck;
int main() {
  long long x;
  cin >> x;
  long long limit = 2 * pow(x, 1.0 / 3);
  long long f1, f2;
  long long res1, res2, res3;
  int cnt = 0;
  for (long long i = 1; i < limit; i++) {
    f1 = (i - 1) * i / 2;
    f2 = (i - 1) * i * (2 * i - 1) / 6;
    res1 = x - f2 + f1 * i;
    res2 = i * i - f1;
    if (res1 % res2 == 0) {
      res3 = res1 / res2;
      if (res3 >= i) {
        fuck.push_back(pair<long long, long long>(i, res3));
        if (res3 == i)
          cnt++;
        else
          cnt += 2;
      }
    }
  }
  cout << cnt << endl;
  int len = fuck.size();
  for (int i = 0; i < len; i++)
    cout << fuck[i].first << " " << fuck[i].second << endl;
  for (int i = len - 1; i >= 0; i--) {
    if (fuck[i].first != fuck[i].second)
      cout << fuck[i].second << " " << fuck[i].first << endl;
  }
}