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>
#pragma comment(linker, "/STACK:20000000")
using namespace std;
int ri() {
  int x;
  scanf("%d", &x);
  return x;
}
long long rll() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void solve() {
  long long x;
  cin >> x;
  set<pair<long long, long long> > result;
  result.insert(make_pair(1, x));
  result.insert(make_pair(x, 1));
  long long tmp = 0;
  for (long long k = 2;; k++) {
    tmp += (k - 1) * (k - 1);
    long long ost = x - tmp;
    long long wtf = k * k * k - (k - 1) * k / 2 * k * 2;
    if (ost < wtf) {
      cerr << k << endl;
      break;
    }
    ost *= 2LL;
    if (ost % k == 0) {
      ost /= k;
      long long n = k;
      long long t = ost + (k - 1) * n;
      long long m = t / (2 * n - k + 1);
      if (m * (2 * n - k + 1) == t) {
        result.insert(make_pair(n, m)), result.insert(make_pair(m, n));
      }
    }
  }
  cout << result.size() << endl;
  for (set<pair<long long, long long> >::iterator it = result.begin();
       it != result.end(); it++)
    cout << it->first << " " << it->second << endl;
}
int main() {
  solve();
  return 0;
}