#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000000;
long long x;
long long ans[maxn][2];
int cnt = 0;
int main() {
  long long i, l, r, cas, f, mid, a2;
  cin >> x;
  for (i = 1;; i++) {
    a2 = x * 6 / i / (i + 1);
    a2 = (a2 + i - 1) / 3;
    if (i < a2 && i * (i + 1) * (3 * a2 - i + 1) / 6 == x) {
      ans[++cnt][0] = i;
      ans[cnt][1] = a2;
    } else if (a2 <= i)
      break;
  }
  cas = cnt;
  f = i;
  cnt *= 2;
  if (i == a2 && (i * (i + 1) * (3 * a2 - i + 1) / 6 == x)) cnt++;
  printf("%d\n", cnt);
  for (i = 1; i <= cas; i++) printf("%I64d %I64d\n", ans[i][0], ans[i][1]);
  if (cnt & 1) printf("%I64d %I64d\n", f, a2);
  for (i = cas; i >= 1; i--) printf("%I64d %I64d\n", ans[i][1], ans[i][0]);
  return 0;
}

### Prompt

Please create a solution in cpp to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000000;
long long x;
long long ans[maxn][2];
int cnt = 0;
int main() {
  long long i, l, r, cas, f, mid, a2;
  cin >> x;
  for (i = 1;; i++) {
    a2 = x * 6 / i / (i + 1);
    a2 = (a2 + i - 1) / 3;
    if (i < a2 && i * (i + 1) * (3 * a2 - i + 1) / 6 == x) {
      ans[++cnt][0] = i;
      ans[cnt][1] = a2;
    } else if (a2 <= i)
      break;
  }
  cas = cnt;
  f = i;
  cnt *= 2;
  if (i == a2 && (i * (i + 1) * (3 * a2 - i + 1) / 6 == x)) cnt++;
  printf("%d\n", cnt);
  for (i = 1; i <= cas; i++) printf("%I64d %I64d\n", ans[i][0], ans[i][1]);
  if (cnt & 1) printf("%I64d %I64d\n", f, a2);
  for (i = cas; i >= 1; i--) printf("%I64d %I64d\n", ans[i][1], ans[i][0]);
  return 0;
}

```