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;
int delta_x[] = {-1, 0, 1, 0};
int delta_y[] = {0, 1, 0, -1};
long long x;
int R;
vector<pair<long long, long long> > r;
inline long long gcd(long long a, long long b) {
  return (a ? gcd(b % a, a) : b);
}
int main() {
  scanf("%I64d", &x);
  long long T;
  long long s = 0, s2 = 0;
  for (long long i = 1; i <= 100000000; i++) {
    s2 += i * i;
    s += i;
    if (x < s2) break;
    if ((x - s2) % s) continue;
    long long M = i + (x - s2) / s;
    r.push_back(pair<long long, long long>(i, M));
    if (i != M) r.push_back(pair<long long, long long>(M, i));
  }
  sort(r.begin(), r.end());
  cout << r.size() << endl;
  int R = r.size() - 1;
  for (int i = (0); i <= (R); i++) {
    printf("%I64d %I64d\n", r[i].first, r[i].second);
  }
}