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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct P {
  long long x;
  long long y;
} ans[1000005];
bool cmp(P k, P kk) { return k.x > kk.x; }
int main() {
  long long x;
  scanf("%I64d", &x);
  long long sum = 0;
  long long dic = 1;
  double fin = pow((double)(7 * x), (double)(1.0 / 3.0));
  for (long long n = 1;; n++) {
    long long tmp = 6 * x + n * n * n - n;
    long long tmp1 = 3 * n * n + 3 * n;
    if (tmp <= 0) break;
    if (tmp / tmp1 >= n && tmp % tmp1 == 0 && tmp / tmp1 == n) {
      sum++;
      ans[dic].x = n;
      ans[dic++].y = tmp / tmp1;
    } else if (tmp / tmp1 >= n && tmp % tmp1 == 0 && tmp / tmp1 != n) {
      sum = sum + 2;
      ans[dic].x = n;
      ans[dic++].y = tmp / tmp1;
      ans[dic].x = ans[dic - 1].y;
      ans[dic++].y = n;
    }
  }
  printf("%I64d\n", sum);
  sort(ans + 1, ans + dic, cmp);
  for (int i = 1; i < dic; i++) {
    printf("%I64d %I64d\n", ans[i].y, ans[i].x);
  }
  return 0;
}