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 int n, i, d[100000], r[100000], sl, da, x, v1, v2, j;
int main() {
  scanf("%lld", &n);
  i = 1;
  sl = 0;
  j = (long long int)sqrt(n);
  if (j > 10000000) j = 10000000;
  while (i <= j) {
    if ((6 * n) % i == 0) {
      x = i;
      v1 = (n - ((x - 1) * x / 2) * (2 * x - 1) / 3 + ((x - 1) * x / 2) * x);
      v2 = (x * x - (x - 1) * x / 2);
      da = v1 / v2;
      if (da >= i && (x * x * da - ((x - 1) * x / 2) * (x + da) +
                      ((x - 1) * x / 2) * (2 * x - 1) / 3) == n) {
        sl++;
        d[sl] = da;
        r[sl] = i;
      }
      if (da < i) break;
    }
    i++;
  }
  if (d[sl] == r[sl])
    printf("%lld\n", 2 * sl - 1);
  else
    printf("%lld\n", 2 * sl);
  for (i = 1; i <= sl; i++) printf("%lld %lld\n", r[i], d[i]);
  if (d[sl] == r[sl]) sl--;
  for (i = sl; i >= 1; i--) printf("%lld %lld\n", d[i], r[i]);
  return 0;
}