#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5;
long long ans[maxn][2];
long long i, n, m, x, la, M, cnt;
long long count(long long n, int &bl) {
  bl = (6 * x + n * n * n - n) % (3 * n * (n + 1));
  return (6 * x + n * n * n - n) / 3 / n / (n + 1);
}
int main() {
  scanf("%lld", &x);
  M = sqrt(6 * x) + 1e-8;
  for (n = 1; n <= M; n++) {
    int bl;
    m = count(n, bl);
    if (m < n) break;
    if (6 * x % n != 0) continue;
    if (bl) continue;
    if (n == m) cnt++;
    la++;
    ans[la][0] = n;
    ans[la][1] = m;
  }
  printf("%lld\n", 2 * la - cnt);
  for (i = 1; i <= la; i++) printf("%lld %lld\n", ans[i][0], ans[i][1]);
  for (i = la; i > 0; i--) {
    if (ans[i][0] == ans[i][1]) continue;
    printf("%lld %lld\n", ans[i][1], ans[i][0]);
  }
  return 0;
}

### Prompt

Generate a cpp solution 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 long long maxn = 1e5;
long long ans[maxn][2];
long long i, n, m, x, la, M, cnt;
long long count(long long n, int &bl) {
  bl = (6 * x + n * n * n - n) % (3 * n * (n + 1));
  return (6 * x + n * n * n - n) / 3 / n / (n + 1);
}
int main() {
  scanf("%lld", &x);
  M = sqrt(6 * x) + 1e-8;
  for (n = 1; n <= M; n++) {
    int bl;
    m = count(n, bl);
    if (m < n) break;
    if (6 * x % n != 0) continue;
    if (bl) continue;
    if (n == m) cnt++;
    la++;
    ans[la][0] = n;
    ans[la][1] = m;
  }
  printf("%lld\n", 2 * la - cnt);
  for (i = 1; i <= la; i++) printf("%lld %lld\n", ans[i][0], ans[i][1]);
  for (i = la; i > 0; i--) {
    if (ans[i][0] == ans[i][1]) continue;
    printf("%lld %lld\n", ans[i][1], ans[i][0]);
  }
  return 0;
}

```