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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long x;
long long sum2(long long a) { return a * (a + 1) * (2 * a + 1) / 6; }
long long sum1(long long a) { return a * (a + 1) / 2; }
long long cnt(long long a, long long b) {
  long long tot = a * b * a;
  tot -= b * sum1(a - 1);
  tot -= a * sum1(a - 1);
  tot += sum2(a - 1);
  return tot;
}
int main() {
  scanf("%lld", &x);
  vector<pair<long long, long long> > ans;
  for (long long i = 1;; i++) {
    if (sum2(i) > x) break;
    long long low = i - 1;
    long long high = 4e18 / i / i;
    while (high - low > 1) {
      if (cnt(i, (low + high + 1) / 2) >= x)
        high = (low + high + 1) / 2;
      else
        low = (low + high + 1) / 2;
    }
    if (cnt(i, high) == x) {
      ans.push_back(make_pair(i, high));
      if (i != high) ans.push_back(make_pair(high, i));
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (auto i : ans) printf("%lld %lld\n", i.first, i.second);
  return 0;
}