#include <bits/stdc++.h>
const int MAXN = 4000005;
using namespace std;
long long x, k;
pair<long long, long long> A[MAXN];
int main() {
  scanf("%lld", &x);
  for (long long n = 1; n <= 2000000; n++) {
    long long m = (6 * x - n + n * n * n);
    long long b = (3 * n * (n + 1));
    if (m > 0 && m % b == 0 && n <= m / b) {
      m /= b;
      A[k++] = {n, m};
      A[k++] = {m, n};
      k -= (n == m);
    }
  }
  sort(A, A + k);
  printf("%lld\n", k);
  for (int i = 0; i < k; i++) printf("%lld %lld\n", A[i].first, A[i].second);
}

### Prompt

Please formulate 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>
const int MAXN = 4000005;
using namespace std;
long long x, k;
pair<long long, long long> A[MAXN];
int main() {
  scanf("%lld", &x);
  for (long long n = 1; n <= 2000000; n++) {
    long long m = (6 * x - n + n * n * n);
    long long b = (3 * n * (n + 1));
    if (m > 0 && m % b == 0 && n <= m / b) {
      m /= b;
      A[k++] = {n, m};
      A[k++] = {m, n};
      k -= (n == m);
    }
  }
  sort(A, A + k);
  printf("%lld\n", k);
  for (int i = 0; i < k; i++) printf("%lld %lld\n", A[i].first, A[i].second);
}

```