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.

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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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> yo[4000006];
int k, ck;
int main() {
  long long ans;
  double y;
  cin >> ans;
  y = 1.000 * ans;
  for (double x = 1; x <= 2000000; x++) {
    y = (2.000 * ans) / (1.00000 * x * (x + 1)) + (x - 1) / 3.000;
    if (y < x) break;
    long long z = (long long)y;
    if (y == z) {
      yo[k] = make_pair((long long)x, z);
      k++;
    }
  }
  if (ans == 1)
    ck = 1, cout << k << endl;
  else if (yo[k - 1].first == yo[k - 1].second) {
    k = 2 * k - 1;
    cout << k << endl;
    k = k / 2 + 1;
    ck = 1;
  } else {
    k = k * 2;
    cout << k << endl;
    k = k / 2;
  }
  for (int i = 0; i < k; i++) {
    printf("%I64d %I64d\n", yo[i].first, yo[i].second);
  }
  for (int i = k - ck - 1; i > -1; i--) {
    printf("%I64d %I64d\n", yo[i].second, yo[i].first);
  }
  return 0;
}