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>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
struct Timer {
  Timer() { beg = clock(); }
  ~Timer() {
    clock_t end = clock();
    cout << "Elapsed time: " << (end - beg + 0.0) / CLOCKS_PER_SEC << "s"
         << endl;
  }
  clock_t beg;
};
int main() {
  long long x;
  cin >> x;
  vector<pair<long long, long long> > ans;
  long long all = x;
  long long k = 1;
  for (; k * k <= x; ++k) {
    x -= (k - 1) * (k - 1);
    if (k * k > 9223372036854775807LL / k) {
      break;
    }
    long long rside = x + k * k * (k - 1) / 2;
    long long denom = (k * k + k) / 2;
    if (rside % denom == 0) {
      long long n = rside / denom;
      ans.push_back(make_pair(n, k));
      if (n != k) {
        ans.push_back(make_pair(k, n));
      }
    }
  }
  sort(ans.begin(), ans.end());
  ans.resize(unique(ans.begin(), ans.end()) - ans.begin());
  cout << ans.size() << endl;
  for (auto &a : ans) {
    printf("%I64d %I64d\n", a.first, a.second);
  }
  return 0;
}