#include <bits/stdc++.h>
using namespace std;
const int DIV = 1000000007;
const double EPS = 0.000001;
long double func1(long double t, long double r, long double v, long double l) {
  return v * t - l - r * sin(v * t / (r * 2)) * 2;
}
long double func2(long double t, long double r, long double v, long double l) {
  return v * t - l + r * sin(v * t / (r * 2)) * 2;
}
int main(int argc, char* argv[]) {
  FILE* f = stdin;
  long long x;
  fscanf(f, "%I64d", &x);
  long long max = 0;
  while (max * max * max < x * 3) max++;
  int k = 0;
  vector<pair<long long, long long>> tab;
  bool kv = false;
  for (long long m = 1; m < max; m++) {
    if ((x * 6) % (m * (m + 1)) == 0) {
      long long tmp = (x * 6) / (m * (m + 1)) + m - 1;
      long long n = tmp / 3;
      if (n * 3 == tmp && m <= n) {
        tab.push_back(make_pair(m, n));
        if (m == n) kv = true;
      }
    }
  }
  printf("%d\n", (int)(2 * tab.size()) - kv);
  for (int i = 0; i < tab.size(); i++) {
    printf("%I64d %I64d\n", tab[i].first, tab[i].second);
  }
  for (int i = tab.size() - 1; i >= 0; i--) {
    if (tab[i].first != tab[i].second) {
      printf("%I64d %I64d\n", tab[i].second, tab[i].first);
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int DIV = 1000000007;
const double EPS = 0.000001;
long double func1(long double t, long double r, long double v, long double l) {
  return v * t - l - r * sin(v * t / (r * 2)) * 2;
}
long double func2(long double t, long double r, long double v, long double l) {
  return v * t - l + r * sin(v * t / (r * 2)) * 2;
}
int main(int argc, char* argv[]) {
  FILE* f = stdin;
  long long x;
  fscanf(f, "%I64d", &x);
  long long max = 0;
  while (max * max * max < x * 3) max++;
  int k = 0;
  vector<pair<long long, long long>> tab;
  bool kv = false;
  for (long long m = 1; m < max; m++) {
    if ((x * 6) % (m * (m + 1)) == 0) {
      long long tmp = (x * 6) / (m * (m + 1)) + m - 1;
      long long n = tmp / 3;
      if (n * 3 == tmp && m <= n) {
        tab.push_back(make_pair(m, n));
        if (m == n) kv = true;
      }
    }
  }
  printf("%d\n", (int)(2 * tab.size()) - kv);
  for (int i = 0; i < tab.size(); i++) {
    printf("%I64d %I64d\n", tab[i].first, tab[i].second);
  }
  for (int i = tab.size() - 1; i >= 0; i--) {
    if (tab[i].first != tab[i].second) {
      printf("%I64d %I64d\n", tab[i].second, tab[i].first);
    }
  }
  return 0;
}

```