#include <bits/stdc++.h>
using namespace std;
long long x, res, n, i, m;
vector<pair<long long, long long> > g;
int main() {
  scanf("%I64d", &x);
  if (x == 1) {
    printf("1\n");
    printf("1 1\n");
    return 0;
  }
  n = 1;
  i = 1;
  m = x / n;
  while (m > i) {
    m = x / n;
    if (!(x % n)) g.push_back(make_pair(i, m));
    i++;
    x += n;
    n += i;
  }
  printf("%d\n", g.size() * 2 -
                     ((g[g.size() - 1].first - g[g.size() - 1].second) == 0));
  for (int i = 0; i < g.size(); i++) {
    printf("%I64d %I64d\n", g[i].first, g[i].second);
  }
  for (int i = g.size() - 1 -
               ((g[g.size() - 1].first - g[g.size() - 1].second) == 0);
       i >= 0; i--) {
    printf("%I64d %I64d\n", g[i].second, g[i].first);
  }
}

### Prompt

Your task is to create 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 x, res, n, i, m;
vector<pair<long long, long long> > g;
int main() {
  scanf("%I64d", &x);
  if (x == 1) {
    printf("1\n");
    printf("1 1\n");
    return 0;
  }
  n = 1;
  i = 1;
  m = x / n;
  while (m > i) {
    m = x / n;
    if (!(x % n)) g.push_back(make_pair(i, m));
    i++;
    x += n;
    n += i;
  }
  printf("%d\n", g.size() * 2 -
                     ((g[g.size() - 1].first - g[g.size() - 1].second) == 0));
  for (int i = 0; i < g.size(); i++) {
    printf("%I64d %I64d\n", g[i].first, g[i].second);
  }
  for (int i = g.size() - 1 -
               ((g[g.size() - 1].first - g[g.size() - 1].second) == 0);
       i >= 0; i--) {
    printf("%I64d %I64d\n", g[i].second, g[i].first);
  }
}

```