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;
template <class T>
void read(T &a) {
  static char c;
  static int fh;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    fh = -1, a = 0;
  else
    fh = 1, a = c - '0';
  while ((c = getchar()) <= '9' && c >= '0') a = (a << 3) + (a << 1) + c - '0';
  a = a * fh;
}
template <class T>
void write(T a) {
  if (a == 0)
    putchar('0');
  else {
    if (a < 0) putchar('-'), a = -a;
    static char c[30];
    static int c0;
    c0 = 0;
    while (a) c[++c0] = a % 10 + '0', a /= 10;
    while (c0) putchar(c[c0--]);
  }
}
long long x, a[2000005][2] = {{0}}, _end, i0;
int tot = 0;
int main() {
  read(x);
  _end = (long long)pow((double)x * 6, 1.0 / 3);
  while (_end * _end * _end < x * 2) _end++;
  for (long long i = 1; i <= _end; i++) {
    i0 = 2 * x + (i * i * i - i) / 3;
    if (i0 % (i * i + i) == 0) {
      i0 /= (i * i + i);
      if (i0 < i) break;
      a[++tot][0] = i;
      a[tot][1] = i0;
    }
  }
  write(tot * 2 - (a[tot][0] == a[tot][1]));
  putchar('\n');
  for (int i = 1; i <= tot; i++)
    write(a[i][0]), putchar(' '), write(a[i][1]), putchar('\n');
  for (int i = (a[tot][0] == a[tot][1]) ? tot - 1 : tot; i >= 1; i--)
    write(a[i][1]), putchar(' '), write(a[i][0]), putchar('\n');
  return 0;
}