#include <bits/stdc++.h>
using namespace std;
long long X;
vector<pair<long long, long long> > V, correctV;
int main() {
  scanf("%I64d", &X);
  X *= 6LL;
  for (int i = 1; 1LL * i * i * i <= X; ++i) {
    long long N = 1LL * i, nw = 1LL * i * (i + 1);
    if (X % nw == 0) {
      long long M = X / nw - 1 + N;
      if (M % 3 != 0) continue;
      M /= 3;
      V.push_back(make_pair(N, M));
      V.push_back(make_pair(M, N));
    }
  }
  for (int i = 1; 1LL * i * i * i <= X; ++i)
    if (X % i == 0) {
      long long third = i, rem = X / i;
      long long Left = 1, Right = 1LL * 3 * 1000 * 1000 * 1000, Mid, Ans = -1;
      while (Left <= Right) {
        Mid = Left + (Right - Left) / 2;
        if (Mid * (Mid + 1) <= rem)
          Ans = Mid, Left = Mid + 1;
        else
          Right = Mid - 1;
      }
      if (Ans * (Ans + 1) == rem) {
        long long N = Ans;
        if ((third - 1 + N) % 3 != 0) continue;
        long long M = (third - 1 + N) / 3;
        V.push_back(make_pair(N, M));
        V.push_back(make_pair(M, N));
      }
    }
  for (int i = 0; i < V.size(); ++i) {
    long long N = V[i].first, M = V[i].second;
    if (N > M) swap(N, M);
    if (N * (N + 1) * (3 * M - N + 1) == X) correctV.push_back(V[i]);
  }
  sort(correctV.begin(), correctV.end());
  int diff = (correctV.size() > 0);
  for (int i = 1; i < correctV.size(); ++i)
    if (correctV[i] != correctV[i - 1]) diff++;
  printf("%i\n", diff);
  if (correctV.size() > 0)
    printf("%I64d %I64d\n", correctV[0].first, correctV[0].second);
  for (int i = 1; i < correctV.size(); ++i)
    if (correctV[i] != correctV[i - 1])
      printf("%I64d %I64d\n", correctV[i].first, correctV[i].second);
}

### 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;
long long X;
vector<pair<long long, long long> > V, correctV;
int main() {
  scanf("%I64d", &X);
  X *= 6LL;
  for (int i = 1; 1LL * i * i * i <= X; ++i) {
    long long N = 1LL * i, nw = 1LL * i * (i + 1);
    if (X % nw == 0) {
      long long M = X / nw - 1 + N;
      if (M % 3 != 0) continue;
      M /= 3;
      V.push_back(make_pair(N, M));
      V.push_back(make_pair(M, N));
    }
  }
  for (int i = 1; 1LL * i * i * i <= X; ++i)
    if (X % i == 0) {
      long long third = i, rem = X / i;
      long long Left = 1, Right = 1LL * 3 * 1000 * 1000 * 1000, Mid, Ans = -1;
      while (Left <= Right) {
        Mid = Left + (Right - Left) / 2;
        if (Mid * (Mid + 1) <= rem)
          Ans = Mid, Left = Mid + 1;
        else
          Right = Mid - 1;
      }
      if (Ans * (Ans + 1) == rem) {
        long long N = Ans;
        if ((third - 1 + N) % 3 != 0) continue;
        long long M = (third - 1 + N) / 3;
        V.push_back(make_pair(N, M));
        V.push_back(make_pair(M, N));
      }
    }
  for (int i = 0; i < V.size(); ++i) {
    long long N = V[i].first, M = V[i].second;
    if (N > M) swap(N, M);
    if (N * (N + 1) * (3 * M - N + 1) == X) correctV.push_back(V[i]);
  }
  sort(correctV.begin(), correctV.end());
  int diff = (correctV.size() > 0);
  for (int i = 1; i < correctV.size(); ++i)
    if (correctV[i] != correctV[i - 1]) diff++;
  printf("%i\n", diff);
  if (correctV.size() > 0)
    printf("%I64d %I64d\n", correctV[0].first, correctV[0].second);
  for (int i = 1; i < correctV.size(); ++i)
    if (correctV[i] != correctV[i - 1])
      printf("%I64d %I64d\n", correctV[i].first, correctV[i].second);
}

```