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.

<image>

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 rdtsc() {
  long long tmp;
  asm("rdtsc" : "=A"(tmp));
  return tmp;
}
inline int myrand() { return abs((rand() << 15) ^ rand()); }
inline int rnd(int x) { return myrand() % x; }
const int INF = (int)1.01e9;
const long double EPS = 1e-9;
void precalc() {}
long long x;
bool read() {
  if (scanf("%lld", &x) < 1) {
    return 0;
  }
  return 1;
}
void solve() {
  vector<pair<long long, long long> > ans;
  for (long long t = 1;; ++t) {
    long long a = t * (t + 1) / 2;
    long long b = a * (2 * t + 1) / 3;
    if (b > x) {
      break;
    }
    long long left = x - b;
    if (left % a) {
      continue;
    }
    left /= a;
    ans.push_back(make_pair(t, t + left));
    if (left) {
      ans.push_back(make_pair(t + left, t));
    }
  }
  printf("%d\n", ((int)(ans).size()));
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ((int)(ans).size()); ++i) {
    printf(
        "%lld"
        " "
        "%lld"
        "\n",
        ans[i].first, ans[i].second);
  }
}
int main() {
  srand(rdtsc());
  precalc();
  while (1) {
    if (!read()) {
      break;
    }
    solve();
  }
  return 0;
}