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;
const long long N = 1e6 + 10;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long n;
struct node {
  long long a, b;
  bool operator<(const node &A) const { return a < A.a; }
} ans[N];
long long len;
signed main() {
  n = read();
  for (long long i = 1; (i * (i + 1) * (2 * i + 1)) / 6 <= n; i++) {
    long long t = (i * (i + 1) * (2 * i + 1)) / 6;
    if (t == n) ans[++len].a = i, ans[len].b = i;
    if (t < n) {
      long long r = i * (i + 1) / 2;
      if ((n - t) % r == 0) {
        ans[++len].a = i;
        ans[len].b = i + ((n - t) / r);
        ans[++len].a = i + ((n - t) / r);
        ans[len].b = i;
      }
    }
  }
  printf("%lld\n", len);
  sort(ans + 1, ans + len + 1);
  for (long long i = 1; i <= len; i++)
    printf("%lld %lld\n", ans[i].a, ans[i].b);
  return 0;
}