#include <bits/stdc++.h>
using namespace std;
long long k;
int nAns = 0, rAns = 0;
long long ansn[1000001], ansm[1000001];
int main() {
  cin >> k;
  for (long long m = 1; m * (m + 1) * (2 * m + 1) <= 6 * k; m++) {
    if ((6 * k + m * (m + 1) * (m - 1)) % (3 * (m + 1) * m) == 0) {
      long long n = (6 * k + m * (m + 1) * (m - 1)) / (3 * (m + 1) * m);
      ansn[nAns] = m;
      ansm[nAns] = n;
      nAns++;
      rAns++;
      if (n != m) {
        rAns++;
      }
    }
  }
  printf("%d\n", rAns);
  for (int i = 0; i < nAns; i++) {
    cout << ansn[i] << " " << ansm[i] << "\n";
  }
  for (int i = nAns - 1; i >= 0; i--) {
    if (ansm[i] == ansn[i]) {
      continue;
    }
    cout << ansm[i] << " " << ansn[i] << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
long long k;
int nAns = 0, rAns = 0;
long long ansn[1000001], ansm[1000001];
int main() {
  cin >> k;
  for (long long m = 1; m * (m + 1) * (2 * m + 1) <= 6 * k; m++) {
    if ((6 * k + m * (m + 1) * (m - 1)) % (3 * (m + 1) * m) == 0) {
      long long n = (6 * k + m * (m + 1) * (m - 1)) / (3 * (m + 1) * m);
      ansn[nAns] = m;
      ansm[nAns] = n;
      nAns++;
      rAns++;
      if (n != m) {
        rAns++;
      }
    }
  }
  printf("%d\n", rAns);
  for (int i = 0; i < nAns; i++) {
    cout << ansn[i] << " " << ansm[i] << "\n";
  }
  for (int i = nAns - 1; i >= 0; i--) {
    if (ansm[i] == ansn[i]) {
      continue;
    }
    cout << ansm[i] << " " << ansn[i] << "\n";
  }
  return 0;
}

```