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>
long long mod = 1e9 + 7;
using namespace std;
long long maxm = 1500000;
int main() {
  int n, m, id;
  long long i;
  long long x;
  cin >> x;
  int ff = 0;
  long long eq;
  vector<pair<long long, long long> > ans;
  ans.clear();
  for (i = 1; i <= maxm; ++i) {
    long long num = 6 * x + i * (i - 1) * (i + 1);
    long long den = 3 * i * (i + 1);
    if (num % den == 0) {
      long long xx = num / den;
      if (xx == i) {
        ff = 1;
        eq = xx;
      } else if (xx > i) {
        ans.push_back(make_pair(i, xx));
      }
    }
  }
  if (ff) {
    int vvv = ans.size();
    cout << vvv * 2 + 1 << endl;
    for (i = 0; i < vvv; ++i)
      cout << ans[i].first << " " << ans[i].second << endl;
    cout << eq << " " << eq << endl;
    for (i = vvv - 1; i >= 0; --i) {
      cout << ans[i].second << " " << ans[i].first << endl;
    }
  } else {
    int vvv = ans.size();
    if (vvv > 0) cout << vvv * 2 << endl;
    for (i = 0; i < vvv; ++i)
      cout << ans[i].first << " " << ans[i].second << endl;
    for (i = vvv - 1; i >= 0; --i) {
      cout << ans[i].second << " " << ans[i].first << endl;
    }
  }
}