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 warning(disable : 4996)
using namespace std;
long long n;
vector<pair<long long, long long> > ans;
int main() {
  long long i, j, k;
  cin >> n;
  for (i = 1; i <= 2000000; i++) {
    long long base = i * (i + 1) / 2 * (2 * i + 1) / 3;
    long long u = i * (i + 1) / 2;
    if (n < base) break;
    if (n == base) {
      ans.push_back(pair<long long, long long>(i, i));
      break;
    }
    if ((n - base) % u == 0) {
      ans.push_back(pair<long long, long long>(i, i + (n - base) / u));
      ans.push_back(pair<long long, long long>(i + (n - base) / u, i));
    }
  }
  sort((ans).begin(), (ans).end());
  printf("%d\n", ans.size());
  for (auto e : ans) printf("%lld %lld\n", e.first, e.second);
  return 0;
}