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.

<image>

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<long long> ans1, ans2;
bool check(long long a, long long c) {
  long long b;
  b = (c + a * (a * (a - 1)) / 2 - (a * (a - 1) * (2 * a - 1)) / 6) %
      (a * a - (a * (a - 1) / 2));
  if (b == 0)
    return 1;
  else
    return 0;
}
bool check2(long long a, long long b, long long c) {
  long long d;
  d = (a * a * b - a * (a * (a - 1) / 2) - b * (a * (a - 1)) / 2 +
       (a * (a - 1) * (2 * a - 1)) / 6);
  if (c == d)
    return 1;
  else
    return 0;
}
bool over(long long a, long long c) {
  long long b, d;
  b = (c + a * (a * (a - 1)) / 2 - (a * (a - 1) * (2 * a - 1)) / 6) /
      (a * a - (a * (a - 1) / 2));
  d = (a * a * b - a * (a * (a - 1) / 2) - b * (a * (a - 1)) / 2 +
       (a * (a - 1) * (2 * a - 1)) / 6);
  if (d > c)
    return 1;
  else
    return 0;
}
bool over2(long long a, long long c) {
  long long b, d;
  b = 1;
  d = (a * a * b - a * (a * (a - 1) / 2) - b * (a * (a - 1)) / 2 +
       (a * (a - 1) * (2 * a - 1)) / 6);
  if (d > c)
    return 1;
  else
    return 0;
}
bool over3(long long a, long long c) {
  long long b, d;
  b = (c + a * (a * (a - 1)) / 2 - (a * (a - 1) * (2 * a - 1)) / 6) /
      (a * a - (a * (a - 1) / 2));
  if (a > b)
    return 1;
  else
    return 0;
}
long long Go(long long a, long long c) {
  long long b, d;
  b = (c + a * (a * (a - 1)) / 2 - (a * (a - 1) * (2 * a - 1)) / 6) /
      (a * a - (a * (a - 1) / 2));
  return b;
}
int main() {
  long long i, j, k, l;
  long long a, b, c, d;
  long long n;
  cin >> n;
  k = sqrt(n) + 1;
  int ans = 0;
  for (i = 1; i <= k; i++) {
    a = i;
    l = Go(a, n);
    if (a * a > n) break;
    if (over(a, n) == 1) break;
    if (over2(a, n) == 1) break;
    if (over3(a, n) == 1) break;
    if (check(a, n) == 1 && check2(a, l, n) == 1) {
      ans1.push_back(a);
      ans2.push_back(l);
      if (a == l) ans--;
    }
  }
  ans += 2 * ans1.size();
  cout << ans << '\n';
  for (i = 0; i < ans1.size(); i++) {
    cout << ans1[i] << ' ' << ans2[i] << '\n';
  }
  for (i--; i >= 0; i--) {
    if (ans1[i] == ans2[i]) continue;
    cout << ans2[i] << ' ' << ans1[i] << '\n';
  }
}