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;
vector<pair<long long, long long> > v;
int main() {
  long long int m = 0, n, i, t = 0, x, ans, val, temp, z = 1e18 + 5, c = 0;
  cin >> x;
  if (x == 1) {
    printf("1\n");
    printf("1 1\n");
    return 0;
  }
  int mx = 10000000;
  while (m < z && mx--) {
    m++;
    if (m == z) break;
    ans = x * 6;
    val = m * (m + 1);
    if (ans % val == 0) {
      temp = (ans / val) + m - 1;
      if (temp % 3 == 0) {
        n = temp / 3;
        z = min(n, z);
        c++;
        if (m == n) t = -1;
        if (m <= n) v.push_back(make_pair(m, n));
      }
    }
  }
  cout << t + v.size() * 2 << endl;
  for (i = 0; i < v.size(); i++) {
    printf("%I64d %I64d\n", v[i].first, v[i].second);
  }
  for (i = v.size() - 1; i >= 0; i--) {
    if (v[i].first == v[i].second) continue;
    printf("%I64d %I64d\n", v[i].second, v[i].first);
  }
}