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;
long long x, tmp;
map<long long, long long> m;
struct NODE {
  long long a, b;
} ans[1000010];
long long ans_len;
long long judge(long long t) {
  if ((x - ((t - 1) * t * (2 * t - 1)) / 6 + ((t - 1) * t * t) / 2) %
          (t * t - (t * (t - 1)) / 2) ==
      0)
    return (x - ((t - 1) * t * (2 * t - 1)) / 6 + ((t - 1) * t * t) / 2) /
           (t * t - (t * (t - 1)) / 2);
  else
    return -1;
}
int main() {
  long long i;
  long long a, b;
  scanf("%I64d", &x);
  ans_len = 0;
  tmp = 0;
  a = 1;
  ans[0].a = 1;
  ans[0].b = x;
  m[x] = 1;
  ans_len++;
  tmp = 0;
  while (1) {
    a++;
    b = judge(a);
    if (a > x / 2) break;
    if (x - ((a - 1) * a * (2 * a - 1)) / 6 < 0) break;
    if (b == -1) continue;
    if (b < a) break;
    ans[ans_len].a = a;
    ans[ans_len++].b = b;
  }
  printf("%I64d\n", ans[ans_len - 1].a == ans[ans_len - 1].b ? ans_len * 2 - 1
                                                             : ans_len * 2);
  for (i = 0; i < ans_len; i++) printf("%I64d %I64d\n", ans[i].a, ans[i].b);
  long long k;
  if (ans[ans_len - 1].a == ans[ans_len - 1].b)
    k = ans_len - 2;
  else
    k = ans_len - 1;
  for (i = k; i >= 0; i--) printf("%I64d %I64d\n", ans[i].b, ans[i].a);
}