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> > ans;
long double f(long double x, long double y) {
  return x * (x + 1) * (y + 1) - (x + y + 2) * (x + 1) * x / 2 +
         x * (x + 1) * (2 * x + 1) / 6;
}
int dcmp(long double x) {
  if (fabs(x) < 1e-10) return 0;
  return x < 0 ? -1 : 1;
}
int main() {
  long double x;
  cin >> x;
  int ret = 0;
  for (int i = 1; i <= 10000000; i++) {
    long double now = f(i, i);
    int flag = dcmp(now - x);
    if (flag > 0) break;
    if (flag == 0) {
      ans.push_back({i, i});
      break;
    }
    long long l = i, r = x + 1;
    while (l + 1 < r) {
      long long mid = (l + r) >> 1;
      if (dcmp(f(i, mid) - x) >= 0)
        r = mid;
      else
        l = mid;
    }
    if (dcmp(f(i, r) - x) == 0) {
      ans.push_back({i, r});
      ans.push_back({r, i});
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  }
  return 0;
}