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;
inline int read() {
  int res = 0;
  int neg;
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9' || ch == '-') {
      if (ch == '-')
        neg = -1;
      else
        neg = 1, res = ch - '0';
      break;
    }
    if (ch == '?') {
      return 2e9 + 1;
    }
  }
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9')
      res *= 10, res += ch - '0';
    else
      break;
  }
  return res * neg;
}
const int maxn = 2020;
const int maxm = 420;
const int MOd = 1e9 + 7;
const int P = 37;
int main() {
  unsigned long long a;
  cin >> a;
  vector<pair<unsigned long long, unsigned long long> > v;
  int maxx = 5000000;
  for (unsigned long long i = 1; i <= maxx; i++) {
    unsigned long long h = i * (i - 1) / 2;
    unsigned long long t = (i + i - 1) * i * (i - 1) / 6;
    unsigned long long s = i * i;
    if (t > a) break;
    if ((a + h * i - t) % (s - h)) continue;
    unsigned long long m = (a + h * i - t) / (s - h);
    if (i >= m) {
      if (i == m)
        v.push_back(pair<unsigned long long, unsigned long long>(i, i));
      break;
    }
    v.push_back(pair<unsigned long long, unsigned long long>(i, m));
    v.push_back(pair<unsigned long long, unsigned long long>(m, i));
  }
  sort(v.begin(), v.end());
  printf("%d\n", v.size());
  for (int i = 0; i < v.size(); i++) {
    printf("%lld %lld\n", v[i].first, v[i].second);
  }
  return 0;
}