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>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <class T>
T sqr(T a) {
  return a * a;
}
template <class T>
T sgn(T a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
int n;
int m;
set<pair<unsigned long long, unsigned long long>> ans;
unsigned long long f(int i, int k) {
  if ((double)(k + 1) * k * (3 * (2 * i + 1) - (2 * k - 1)) / 12 > 5e18)
    return 2e18;
  return (unsigned long long)(k + 1) * k *
         ((unsigned long long)3 * (2 * i + 1) - (2 * k + 1)) / 12;
}
int main() {
  unsigned long long x;
  cin >> x;
  for (int i = 1; i < 2000000; i++) {
    unsigned long long res = 0;
    int l = 0, r = i;
    while (r - l > 1) {
      int m = (r + l) / 2;
      if (f(i, m) < x)
        l = m;
      else
        r = m;
    }
    res = f(i, r);
    int j = r;
    if (res == x) {
      if (ans.count(make_pair(i, j))) break;
      ans.insert(make_pair(i, j));
      ans.insert(make_pair(j, i));
    } else if (res < x && j == i) {
      unsigned long long r = (unsigned long long)i * (i + 1) / 2;
      unsigned long long w = j + (x - res) / r;
      if ((x - res) % r == 0) {
        if (ans.count(make_pair(i, w))) break;
        ans.insert(make_pair(i, w));
        ans.insert(make_pair(w, i));
      }
    }
  }
  cout << ans.size() << endl;
  for (auto cur : ans) cout << cur.first << ' ' << cur.second << endl;
  return 0;
}