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>
const double pi = acos(-1);
const int inf = INT_MAX;
const int N = 1E7 + 99;
long long a[N], b[N];
int n;
int main() {
  long long y, l1, l2, l3, l4;
  while (scanf("%I64d", &y) != EOF) {
    l1 = 1;
    n = 0;
    bool f = 0;
    while (l1 <= y) {
      l2 = l1 * (l1 - 1) * (2 * l1 - 1) / 6;
      l3 = l1 * (l1 - 1) * l1 / 2;
      l2 = y - l2 + l3;
      if (l2 <= 0) break;
      l3 = l1 * l1 - l1 * (l1 - 1) / 2;
      if (l2 % l3) {
        l1++;
        continue;
      }
      l2 /= l3;
      if (l2 < l1) break;
      a[n] = l1, b[n++] = l2;
      if (l1 == l2) f = 1;
      l1++;
    }
    if (f)
      printf("%d\n", 2 * n - 1);
    else
      printf("%d\n", 2 * n);
    for (int i = 0; i < n; i++) printf("%I64d %I64d\n", a[i], b[i]);
    for (int i = n - 1; i >= 0; i--)
      if (a[i] != b[i]) printf("%I64d %I64d\n", b[i], a[i]);
  }
  return 0;
}