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;
vector<pair<long long, long long> > answer;
long long ara[1000005];
int main() {
  long long p, i = 1;
  cin >> p;
  for (i = 1; i < 10000005; i++) {
    long long temp = (i * (i + 1) * (2ll * i + 1)) / 6;
    if (temp > p) break;
    temp = p - temp;
    long long xx = (i * (i + 1)) / 2;
    if (temp % (xx) == 0) answer.push_back(make_pair(i, i + (temp / (xx))));
  }
  int total = answer.size();
  if (!total) {
    printf("0\n");
    return 0;
  }
  total *= 2;
  if (answer[answer.size() - 1].first == answer[answer.size() - 1].second)
    total--;
  printf("%d\n", total);
  for (i = 0; i < answer.size(); i++)
    printf("%I64d %I64d\n", answer[i].first, answer[i].second);
  for (i = answer.size() - 1; i >= 0; i--) {
    if (answer[i].first == answer[i].second) continue;
    printf("%I64d %I64d\n", answer[i].second, answer[i].first);
  }
  return 0;
}