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;
long long x, l, r, mid, k, sz;
vector<pair<long long, long long> > V;
long long f(long long n, long long m) {
  return n * (n + 1) * (3 * m + 1 - n) / 6;
}
int main() {
  scanf("%I64d", &k);
  x = 1;
  sz = k;
  while (f(x, x) <= k) {
    l = x;
    r = sz;
    while (l < r) {
      mid = (l + r) >> 1;
      if (f(x, mid) < k)
        l = mid + 1;
      else
        r = mid;
    }
    if (f(x, r) == k) V.push_back({x, r});
    sz = r;
    ++x;
  }
  sz = V.size();
  if (V[sz - 1].first == V[sz - 1].second) {
    printf("%I64d\n", sz * 2 - 1);
    for (int i = 0; i < sz; ++i)
      printf("%I64d %I64d\n", V[i].first, V[i].second);
    for (int i = sz - 2; i >= 0; --i)
      printf("%I64d %I64d\n", V[i].second, V[i].first);
  } else {
    printf("%I64d\n", sz << 1);
    for (int i = 0; i < sz; ++i)
      printf("%I64d %I64d\n", V[i].first, V[i].second);
    for (int i = sz - 1; i >= 0; --i)
      printf("%I64d %I64d\n", V[i].second, V[i].first);
  }
  return 0;
}