#include <bits/stdc++.h>
using namespace std;
long long x;
set<pair<long long, long long> > answ;
int main() {
  scanf("%I64d", &x);
  long long NM, N, M, EMPTY;
  long long top, down, parb;
  NM = 0;
  N = M = EMPTY = 0;
  for (long long n = 1;; n++) {
    NM = n;
    N += (n)-1;
    M += (n)-1;
    EMPTY += (n - 1) * (n - 1);
    parb = NM * (n * n) - N * n - M * n + EMPTY;
    if (parb > x) break;
    top = (x + N * n - EMPTY);
    down = (NM * n - M);
    if (top % down == 0) {
      long long m = top / down;
      answ.insert(make_pair(n, m));
      answ.insert(make_pair(m, n));
      if (m < n) break;
    }
  }
  printf("%d\n", (int)answ.size());
  for (set<pair<long long, long long> >::iterator it = answ.begin();
       it != answ.end(); it++) {
    printf("%I64d %I64d\n", it->first, it->second);
  }
}

### 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;
set<pair<long long, long long> > answ;
int main() {
  scanf("%I64d", &x);
  long long NM, N, M, EMPTY;
  long long top, down, parb;
  NM = 0;
  N = M = EMPTY = 0;
  for (long long n = 1;; n++) {
    NM = n;
    N += (n)-1;
    M += (n)-1;
    EMPTY += (n - 1) * (n - 1);
    parb = NM * (n * n) - N * n - M * n + EMPTY;
    if (parb > x) break;
    top = (x + N * n - EMPTY);
    down = (NM * n - M);
    if (top % down == 0) {
      long long m = top / down;
      answ.insert(make_pair(n, m));
      answ.insert(make_pair(m, n));
      if (m < n) break;
    }
  }
  printf("%d\n", (int)answ.size());
  for (set<pair<long long, long long> >::iterator it = answ.begin();
       it != answ.end(); it++) {
    printf("%I64d %I64d\n", it->first, it->second);
  }
}

```