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<long long> ns;
vector<long long> ms;
int main() {
  long long x, t, c, a, n, m;
  cin >> x;
  int k = 0;
  int equ = 0;
  for (n = 1;; n++) {
    t = n * (n + 1) / 2;
    c = (n) * (n + 1) * (2 * n + 1) / 6 - n * t;
    if (c > x) break;
    a = n * (n + 1) - t;
    if ((x - c) % a == 0) {
      m = (x - c) / a;
      if (m < n) break;
      ns.push_back(n);
      ms.push_back(m);
      k++;
      if (m == n) {
        equ = 1;
        break;
      }
    }
  }
  if (equ) {
    k = 2 * k - 1;
    printf("%d\n", k);
    int sz = ns.size();
    for (int i = 0; i < sz; i++) {
      printf("%I64d %I64d\n", ns[i], ms[i]);
    }
    for (int i = sz - 2; i >= 0; i--) {
      printf("%I64d %I64d\n", ms[i], ns[i]);
    }
  } else {
    k = 2 * k;
    printf("%d\n", k);
    int sz = ns.size();
    for (int i = 0; i < sz; i++) {
      printf("%I64d %I64d\n", ns[i], ms[i]);
    }
    for (int i = sz - 1; i >= 0; i--) {
      printf("%I64d %I64d\n", ms[i], ns[i]);
    }
  }
  return 0;
}