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;
const long long N = 1e5 + 5, mod = 1e9 + 7, inf = 1e18 + 7;
vector<pair<long long, long long> > b;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long x;
  cin >> x;
  for (long long i = 1; i * i * i <= 8 * x; i++) {
    if ((6 * x) % (i * (i + 1))) {
      continue;
    }
    long long a = (6 * x) / (i * (i + 1));
    a = a + i - 1;
    if (a % 3) {
      continue;
    }
    if (i > a / 3) {
      break;
    }
    b.push_back({i, a / 3});
  }
  if (b[b.size() - 1].first == b[b.size() - 1].second) {
    cout << 2 * b.size() - 1 << '\n';
  } else {
    cout << 2 * b.size() << '\n';
  }
  for (long long i = 0; i < b.size(); i++) {
    if (b[i].first == b[i].second) {
      continue;
    }
    cout << b[i].first << " " << b[i].second << '\n';
  }
  for (long long i = b.size() - 1; i >= 0; i--) {
    cout << b[i].second << " " << b[i].first << '\n';
  }
}