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> > v, v1;
int main() {
  long long x;
  scanf("%lld", &x);
  long long X = x * 6;
  for (long long i = 1; i <= 1500000; i++) {
    long long tmp = i * (i + 1);
    if (X % tmp != 0) continue;
    long long y = X / tmp;
    y += i;
    y--;
    if (y % 3 != 0) continue;
    y /= 3;
    if (y < i) break;
    v.push_back(make_pair(i, y));
  }
  for (int i = v.size() - 1; i >= 0; i--) {
    if (v[i].first != v[i].second)
      v1.push_back(make_pair(v[i].second, v[i].first));
  }
  int sz = v.size() + v1.size();
  printf("%d\n", sz);
  for (int i = 0; i < v.size(); i++)
    printf("%lld %lld\n", v[i].first, v[i].second);
  for (int i = 0; i < v1.size(); i++)
    printf("%lld %lld\n", v1[i].first, v1[i].second);
  return 0;
}