#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> fun(long long x, long long n) {
  long long i = 0, j = 0, k = 0, tmp1 = 0, tmp2 = 0, tmp3 = 0, tmp = 0,
            tmp4 = 0, tmp5 = 0, flag = 0, T = 0, N = 0, Q = 0;
  tmp = 6 * x - n + n * n * n;
  if (tmp == 0) return make_pair(-1, -1);
  if (tmp % 3) return make_pair(-1, -1);
  tmp /= 3;
  if (tmp % n) return make_pair(-1, -1);
  tmp /= n;
  if (tmp % (n + 1)) return make_pair(-1, -1);
  tmp /= (n + 1);
  if (n > tmp) return make_pair(-1, -1);
  return make_pair(n, tmp);
}
int main() {
  long long i = 0, j = 0, k = 0, tmp1 = 0, tmp2 = 0, tmp3 = 0, tmp = 0,
            tmp4 = 0, tmp5 = 0, flag = 0, T = 0, N = 0, Q = 0;
  long long x;
  pair<long long, long long> pr;
  cin >> x;
  vector<pair<long long, long long> > ans;
  tmp = 10000001;
  for (i = 1; i < min(x + 1, tmp); i++) {
    pr = fun(x, i);
    if (pr.first != -1) ans.push_back(pr);
  }
  if (ans.size() == 0) {
    printf("Empty");
    return 0;
  }
  if (ans.back().first == ans.back().second) {
    printf("%d\n", 2 * ans.size() - 1);
    for (i = 0; i < ans.size(); i++) {
      cout << ans[i].first << " " << ans[i].second << endl;
    }
    if (ans.size() > 1)
      for (i = ans.size() - 2; i > -1; i--) {
        cout << ans[i].second << " " << ans[i].first << endl;
      }
  } else {
    printf("%d\n", 2 * ans.size());
    for (i = 0; i < ans.size(); i++) {
      cout << ans[i].first << " " << ans[i].second << endl;
    }
    for (i = ans.size() - 1; i > -1; i--) {
      cout << ans[i].second << " " << ans[i].first << endl;
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
pair<long long, long long> fun(long long x, long long n) {
  long long i = 0, j = 0, k = 0, tmp1 = 0, tmp2 = 0, tmp3 = 0, tmp = 0,
            tmp4 = 0, tmp5 = 0, flag = 0, T = 0, N = 0, Q = 0;
  tmp = 6 * x - n + n * n * n;
  if (tmp == 0) return make_pair(-1, -1);
  if (tmp % 3) return make_pair(-1, -1);
  tmp /= 3;
  if (tmp % n) return make_pair(-1, -1);
  tmp /= n;
  if (tmp % (n + 1)) return make_pair(-1, -1);
  tmp /= (n + 1);
  if (n > tmp) return make_pair(-1, -1);
  return make_pair(n, tmp);
}
int main() {
  long long i = 0, j = 0, k = 0, tmp1 = 0, tmp2 = 0, tmp3 = 0, tmp = 0,
            tmp4 = 0, tmp5 = 0, flag = 0, T = 0, N = 0, Q = 0;
  long long x;
  pair<long long, long long> pr;
  cin >> x;
  vector<pair<long long, long long> > ans;
  tmp = 10000001;
  for (i = 1; i < min(x + 1, tmp); i++) {
    pr = fun(x, i);
    if (pr.first != -1) ans.push_back(pr);
  }
  if (ans.size() == 0) {
    printf("Empty");
    return 0;
  }
  if (ans.back().first == ans.back().second) {
    printf("%d\n", 2 * ans.size() - 1);
    for (i = 0; i < ans.size(); i++) {
      cout << ans[i].first << " " << ans[i].second << endl;
    }
    if (ans.size() > 1)
      for (i = ans.size() - 2; i > -1; i--) {
        cout << ans[i].second << " " << ans[i].first << endl;
      }
  } else {
    printf("%d\n", 2 * ans.size());
    for (i = 0; i < ans.size(); i++) {
      cout << ans[i].first << " " << ans[i].second << endl;
    }
    for (i = ans.size() - 1; i > -1; i--) {
      cout << ans[i].second << " " << ans[i].first << endl;
    }
  }
  return 0;
}

```