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;
const unsigned long long siz = 5000005;
unsigned long long sum[siz];
int main() {
  unsigned long long x;
  cin >> x;
  sum[1] = 0;
  for (int i = 2; i < siz; i++) {
    sum[i] = sum[i - 1] + (i - 1);
  }
  vector<pair<unsigned long long, unsigned long long>> arr(siz, {0, 0});
  unsigned long long i = 1;
  unsigned long long counter = 0;
  unsigned long long n = 0;
  unsigned long long cur_sum = 0;
  do {
    cur_sum += sum[i];
    unsigned long long calc = (x + cur_sum) * 2 % (i * (i + 1));
    if (calc == 0) {
      n = (x + cur_sum) * 2 / (i * (i + 1));
      if (i <= n) {
        arr[counter] = {i, n};
        counter++;
        i++;
      }
    } else {
      i++;
    }
  } while (i <= n && i < siz);
  int add = 0;
  if (arr[counter - 1].first == arr[counter - 1].second) {
    add++;
  }
  cout << counter * 2 - add << endl;
  for (int i = 0; i < counter; i++) {
    cout << arr[i].first << " " << arr[i].second << endl;
  }
  for (int i = counter - 1 - add; i >= 0; i--) {
    cout << arr[i].second << " " << arr[i].first << endl;
  }
}