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;
bool vis[10001010];
vector<pair<long long, long long> > box;
int main() {
  long long int n;
  cin >> n;
  for (int i = 1; i <= min(n, (long long)2000000); i++) {
    if (vis[i]) continue;
    long long int m = i;
    long long v;
    long long int div;
    long long ans;
    v = m * (m - 1);
    v = v / 2;
    v = v * m;
    v = v * -1;
    v += ((m * (m - 1) * (2 * m - 1)) / 6);
    v -= n;
    div = ((m * (m - 1)) / 2 - m * m);
    if (div == 0 || v % div != 0) continue;
    ans = v / div;
    if (ans >= m && ans <= 1000000000000000000) {
      vis[i] = 1;
      box.push_back(make_pair(m, ans));
      if (m != ans) box.push_back(make_pair(ans, m));
    } else if (ans > 1000000000000000000)
      break;
  }
  cout << box.size() << endl;
  sort(box.begin(), box.end());
  for (int i = 0; i < box.size(); i++) {
    cout << box[i].first << " " << box[i].second << endl;
  }
}