#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 500;
long long int x, ou1[maxn], ou2[maxn];
int main() {
  cin >> x;
  long long int n, u = 6LL * x;
  int tot = 0;
  for (n = 1LL;; n++) {
    long long int p =
        u + 3LL * n * n * (n - 1LL) - n * (n - 1LL) * (2LL * n - 1LL);
    long long int t = 3LL * n * n + 3LL * n;
    if (p / t < n) break;
    if (p % t) continue;
    long long int m = p / t;
    ou1[++tot] = n, ou2[tot] = m;
  }
  int i;
  if (ou1[tot] == ou2[tot])
    cout << 2 * tot - 1 << endl;
  else
    cout << 2 * tot << endl;
  for (i = 1; i <= tot; i++) cout << ou1[i] << ' ' << ou2[i] << endl;
  if (ou1[tot] == ou2[tot]) tot--;
  for (i = tot; i > 0; i--) cout << ou2[i] << ' ' << ou1[i] << endl;
}

### Prompt

Please create a solution in CPP 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;
const int maxn = 1e6 + 500;
long long int x, ou1[maxn], ou2[maxn];
int main() {
  cin >> x;
  long long int n, u = 6LL * x;
  int tot = 0;
  for (n = 1LL;; n++) {
    long long int p =
        u + 3LL * n * n * (n - 1LL) - n * (n - 1LL) * (2LL * n - 1LL);
    long long int t = 3LL * n * n + 3LL * n;
    if (p / t < n) break;
    if (p % t) continue;
    long long int m = p / t;
    ou1[++tot] = n, ou2[tot] = m;
  }
  int i;
  if (ou1[tot] == ou2[tot])
    cout << 2 * tot - 1 << endl;
  else
    cout << 2 * tot << endl;
  for (i = 1; i <= tot; i++) cout << ou1[i] << ' ' << ou2[i] << endl;
  if (ou1[tot] == ou2[tot]) tot--;
  for (i = tot; i > 0; i--) cout << ou2[i] << ' ' << ou1[i] << endl;
}

```