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> > ans;
int main() {
  long long x, m, n, y;
  cin >> x;
  ans.clear();
  ans.push_back(make_pair((long long)1, x));
  long long b = 1, e = 1500000, md;
  while (b <= e) {
    md = (b + e) >> 1;
    if (((md + 2) * (md + 1) * (2 * md + 3)) / 6 > x &&
        (md * (md + 1) * (2 * md + 1)) / 6 <= x)
      break;
    else if ((md * (md + 1) * (2 * md + 1)) / 6 <= x)
      b = md + 1;
    else
      e = md - 1;
  }
  for (long long i = 2; i < md + 1; i++) {
    y = x + (i * i * (i - 1)) / 2 - (i * (i - 1) * (2 * i - 1)) / 6;
    n = i * i - (i * (i - 1)) / 2;
    if (y % n == 0) ans.push_back(make_pair(i, y / n));
  }
  n = ans.size();
  if (ans[n - 1].first != ans[n - 1].second) {
    m = 1;
    cout << 2 * ans.size() << endl;
  } else {
    m = 2;
    cout << 2 * n - 1 << endl;
  }
  for (typeof(ans.begin()) it = ans.begin(); it != ans.end(); it++)
    cout << (*it).first << " " << (*it).second << endl;
  for (long long i = m; i < ans.size() + 1; i++)
    cout << ans[n - i].second << " " << ans[n - i].first << endl;
}