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;
struct sb {
  long long x;
  long long y;
  bool operator<(const sb &wakaka) const { return x < wakaka.x; }
} test;
set<sb> node;
int main() {
  long long x;
  cin >> x;
  long long len = 2 * pow((double)x, (double)1 / (double)3);
  long long cnt = 0;
  long long n = 0;
  while (1) {
    n++;
    if (n > len) break;
    long long en = x * 6;
    long long i2 = (n - 1) * n * (2 * n - 1);
    en = en - i2;
    long long he = (n - 1) * n * 3;
    en = en + he * n;
    long long la = 6 * n * n - he;
    if (en % la == 0) {
      long long ans = en / la;
      test.x = n;
      test.y = ans;
      if (ans < n) break;
      node.insert(test);
      test.y = n;
      test.x = ans;
      node.insert(test);
      cnt = cnt + 2;
    }
  }
  cout << node.size() << endl;
  for (set<sb>::iterator it = node.begin(); it != node.end(); it++) {
    test = *it;
    cout << test.x << " " << test.y << endl;
  }
  return 0;
}