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;
vector<pair<long long, long long> > ans;
bool cmp(pair<long long, long long> A, pair<long long, long long> B) {
  return A.first < B.first || A.first == B.first && A.second < B.second;
}
int main() {
  long long C, n, m, x, A, B;
  scanf("%I64d", &x);
  n = 1;
  C = 0;
  while (1) {
    C += n * n;
    A = x - C;
    if (A < 0) break;
    A = A * 2;
    B = n * (n + 1);
    if (A % B == 0) {
      B = A / B;
      if (n + B > 0) {
        m = n + B;
        ans.push_back(make_pair(n, m));
        if (n != m) ans.push_back(make_pair(m, n));
      }
    }
    n++;
  }
  n = ans.size();
  printf("%I64d\n", n);
  sort(ans.begin(), ans.end(), cmp);
  for (long long i = 0; i < n; i++) {
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  }
}