#include <bits/stdc++.h>
using namespace std;
long long ans1[1000000], ans2[1000000];
struct node {
  long long a, b;
} ans[1000000];
int cmp(node a, node b) { return a.a < b.a; }
int main() {
  long long sum = 0, a, b;
  cin >> a;
  for (long long i = 1; i <= 4e7; i++) {
    long long n = 6 * a + i * i * i - i;
    long long m = 3 * i * (i + 1);
    if (n % m == 0) {
      if (i > n / m) break;
      ans[sum++].a = i;
      ans[sum - 1].b = n / m;
      if (i != n / m) {
        ans[sum++].a = n / m;
        ans[sum - 1].b = i;
      }
    }
  }
  sort(ans, ans + sum, cmp);
  cout << sum << endl;
  for (int i = 0; i < sum; i++) cout << ans[i].a << ' ' << ans[i].b << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
long long ans1[1000000], ans2[1000000];
struct node {
  long long a, b;
} ans[1000000];
int cmp(node a, node b) { return a.a < b.a; }
int main() {
  long long sum = 0, a, b;
  cin >> a;
  for (long long i = 1; i <= 4e7; i++) {
    long long n = 6 * a + i * i * i - i;
    long long m = 3 * i * (i + 1);
    if (n % m == 0) {
      if (i > n / m) break;
      ans[sum++].a = i;
      ans[sum - 1].b = n / m;
      if (i != n / m) {
        ans[sum++].a = n / m;
        ans[sum - 1].b = i;
      }
    }
  }
  sort(ans, ans + sum, cmp);
  cout << sum << endl;
  for (int i = 0; i < sum; i++) cout << ans[i].a << ' ' << ans[i].b << endl;
}

```