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;
template <class T>
inline T sqr(const T& a) {
  return a * a;
}
const long double EPS = 1e-9;
const long double PI = 2 * acos(0.0);
const int N = 100;
long long Calc(pair<long long, long long> p) {
  long long n = p.first, m = p.second;
  if (n < m) swap(n, m);
  long long r = 0;
  for (long long x = m; x >= 1; --x) {
    r += (m - x + 1) * (n - x + 1);
  }
  return r;
}
void Check(long long x, vector<pair<long long, long long> >& s) {
  for (pair<long long, long long> p : s) {
    assert(Calc(p) == x);
    if (Calc(p) != x) {
      cout << "! " << p.first << " " << p.second << endl;
    }
  }
}
int main() {
  long long x;
  cin >> x;
  vector<pair<long long, long long> > s;
  for (long long i = 1; i <= 1500000; ++i) {
    long long tg = x * 3;
    long long d = i * (i + 1) / 2;
    if (tg % d == 0) {
      tg /= d;
      long long b = 2 * i + 1;
      if (tg < b) break;
      tg -= b;
      if (tg % 3 == 0) {
        tg /= 3;
        long long j = i + tg;
        s.push_back({i, j});
        s.push_back({j, i});
      }
    }
  }
  sort(s.begin(), s.end());
  s.erase(unique(s.begin(), s.end()), s.end());
  int n = int(s.size());
  printf("%d\n", n);
  for (int i = 0; i < n; ++i) {
    cout << s[i].first << ' ' << s[i].second << '\n';
  }
  return 0;
}