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;
using ll = long long;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ll x;
  cin >> x;
  auto find_sol = [&](ll n) -> ll {
    ll a1 = n * n;
    ll c1 = n * n * (n - 1);
    c1 /= 2;
    ll a2 = n * (n - 1);
    a2 /= 2;
    ll c2 = (n - 1) * n * (2 * n - 1);
    c2 /= 6;
    if (a1 <= a2) return -1;
    ll m = (x - c2 + c1) / (a1 - a2);
    return m;
  };
  vector<pair<ll, ll>> vp;
  for (ll i = 1; i <= 1e7; i++) {
    ll m = find_sol(i);
    if (m < i) continue;
    ll n = i;
    if (n * n * m - (n + m) * (n * (n - 1)) / 2 +
            ((n - 1) * n * (2 * n - 1)) / 6 !=
        x)
      continue;
    vp.push_back(make_pair(i, m));
    if (m != i) vp.push_back(make_pair(m, i));
  }
  sort(vp.begin(), vp.end());
  cout << vp.size() << "\n";
  for (auto p : vp) {
    cout << p.first << " " << p.second << "\n";
  }
  return 0;
}