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;
map<long long, long long> mp;
int main() {
  long long x = 0;
  long long n = 0;
  scanf("%lld\n", &x);
  if (x == 1) {
    printf("%d\n%d %d\n", 1, 1, 1);
    return 0;
  }
  for (long long i = 1; i <= x; ++i) {
    long long a = (i * (i + 1LL)) / 2LL;
    long long b =
        x + (i * i * (i - 1LL)) / 2LL - (i * (i - 1) * (2LL * i - 1LL)) / 6LL;
    long long d = b / a;
    if (i > d) {
      n *= 2;
      break;
    }
    if (b % a == 0) {
      long long c = b / a;
      mp[c] = i;
      ++n;
      if (i == c) {
        n *= 2;
        --n;
        break;
      }
    }
  }
  printf("%lld\n", n);
  map<long long, long long>::iterator it = mp.end();
  --it;
  for (;; --it) {
    if (it->first == it->second) {
      if (it == mp.begin()) break;
      continue;
    }
    printf("%lld %lld\n", it->second, it->first);
    if (it == mp.begin()) break;
  }
  it = mp.begin();
  for (; it != mp.end(); ++it) {
    printf("%lld %lld\n", it->first, it->second);
  }
  return 0;
}