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 int, long long int> > res;
int main() {
  long long int x, c, i;
  scanf("%I64d", &x);
  for (i = 1LL; i * (i + 1LL) * (2LL * i + 1LL) / 6LL <= x; i++) {
    c = i * (i + 1LL) * (2LL * i + 1LL) / 6LL;
    if ((x - c) % (i * (i + 1LL) / 2LL) == 0LL) {
      res.push_back(make_pair(i, i + (x - c) / (i * (i + 1LL) / 2LL)));
    }
  }
  if (res.back().first == res.back().second) {
    printf("%d\n", res.size() * 2 - 1);
    for (i = 0; i < res.size(); i++) {
      printf("%I64d %I64d\n", res[i].first, res[i].second);
    }
    for (i = res.size() - 2LL; i >= 0LL; i--) {
      printf("%I64d %I64d\n", res[i].second, res[i].first);
    }
  } else {
    printf("%d\n", res.size() * 2);
    for (i = 0; i < res.size(); i++) {
      printf("%I64d %I64d\n", res[i].first, res[i].second);
    }
    for (i = res.size() - 1LL; i >= 0LL; i--) {
      printf("%I64d %I64d\n", res[i].second, res[i].first);
    }
  }
}