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;
int main(void) {
  long long x;
  scanf("%lld", &x);
  long long n[10000], m[10000];
  long long t1, t2;
  int count = 0;
  for (long long i = 1; i <= 1e7; i++) {
    t1 = 6 * x + i * i * i - i;
    t2 = 3 * i * i + 3 * i;
    if (t1 % t2 != 0) continue;
    t1 = t1 / t2;
    if (i <= t1) {
      n[count] = i;
      m[count] = t1;
      count++;
      if (i == t1) break;
    } else
      break;
  }
  if (n[count - 1] == m[count - 1])
    printf("%d\n", 2 * count - 1);
  else
    printf("%d\n", 2 * count);
  for (int i = 0; i < count; i++) printf("%lld %lld\n", n[i], m[i]);
  if (n[count - 1] == m[count - 1]) count--;
  for (int i = count - 1; i >= 0; i--) printf("%lld %lld\n", m[i], n[i]);
  return 0;
}