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> > ans;
long long squ_sum(long long i) {
  long long ans = (i + 1) * i * (2 * i + 1) / 6;
  return ans;
}
long long sum(long long i) { return i * (i + 1) / 2; }
long long _f(long long l, long long x) {
  long long son = (x - squ_sum(l - 1) + l * sum(l - 1));
  long long mum = l * l - sum(l - 1);
  return son / mum;
}
long long f(long long l, long long r) {
  long long ans = l * l * r - (l + r) * sum(l - 1) + squ_sum(l - 1);
  return ans;
}
void solve(long long x) {
  for (long long l = 1; l <= x; l++) {
    long long r = _f(l, x);
    if (f(l, r) == x && l <= r) {
      ans.push_back(make_pair(l, r));
    }
    if (r <= l) break;
  }
  vector<pair<long long, long long> > tot;
  for (auto i : ans) {
    if (i.first == i.second)
      tot.push_back(i);
    else {
      tot.push_back(i);
      tot.push_back(make_pair(i.second, i.first));
    }
  }
  sort(tot.begin(), tot.end());
  cout << tot.size() << endl;
  for (auto i : tot) cout << i.first << " " << i.second << endl;
}
int main(int argc, const char* argv[]) {
  long long x;
  cin >> x;
  solve(x);
  return 0;
}