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(0);
  cin.tie(0);
  ll x;
  cin >> x;
  int k = 0;
  for (ll n = 1; n <= (int)1.5e6; ++n) {
    ll a = 6 * x + n * (n - 1) * (n + 1), b = 3 * n * (n + 1);
    if (a % b == 0 && a / b >= n) {
      ++k;
    }
  }
  for (ll n = (int)1.5e6; n >= 1; --n) {
    ll a = 6 * x + n * (n - 1) * (n + 1), b = 3 * n * (n + 1);
    if (a % b == 0 && a / b > n) {
      ++k;
    }
  }
  cout << k << endl;
  for (ll n = 1; n <= (int)1.5e6; ++n) {
    ll a = 6 * x + n * (n - 1) * (n + 1), b = 3 * n * (n + 1);
    if (a % b == 0 && a / b >= n) {
      cout << n << ' ' << a / b << endl;
    }
  }
  for (ll n = (int)1.5e6; n >= 1; --n) {
    ll a = 6 * x + n * (n - 1) * (n + 1), b = 3 * n * (n + 1);
    if (a % b == 0 && a / b > n) {
      cout << a / b << ' ' << n << endl;
    }
  }
}