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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<unsigned long long, unsigned long long> > v;
unsigned long long N = 2e6;
unsigned long long getn(unsigned long long n, unsigned long long x) {
  unsigned long long ret = 6 * x;
  ret += (n * n * n);
  ret -= n;
  return ret;
}
unsigned long long getd(unsigned long long n) {
  unsigned long long ret = n * n;
  ret += n;
  ret *= 3;
  return ret;
}
int main() {
  unsigned long long x;
  cin >> x;
  unsigned long long ans = 0;
  for (int n = 1; n <= N; n++) {
    unsigned long long num = getn(n, x);
    unsigned long long den = getd(n);
    if (num % den) continue;
    unsigned long long m = num / den;
    if (n > m) continue;
    ans++;
    if (n != m) ans++;
    v.emplace_back(make_pair(n, m));
  }
  cout << ans << endl;
  for (auto it : v) {
    cout << it.first << " " << it.second << endl;
  }
  reverse(v.begin(), v.end());
  for (auto it : v) {
    if (it.first != it.second) cout << it.second << " " << it.first << endl;
  }
}