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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int read() {
  char c;
  int ans = 0, f = 1;
  c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans * f;
}
long long f(long long i) { return i * (i + 1) * (2 * i + 1) / 6; }
long long a[1000006], b[1000006];
int main() {
  long long n;
  int cnt = 0, ans = 0;
  scanf("%I64d", &n);
  for (long long i = 1; i * i <= n && f(i) <= n; i++) {
    long long fmm = f(i);
    long long ne = i * (i + 1) / 2;
    if ((n - fmm) % ne == 0) {
      a[cnt] = i;
      b[cnt++] = i + (n - fmm) / ne;
      if (n - fmm == 0) ans = -1;
    }
  }
  ans += cnt * 2;
  printf("%d\n", ans);
  for (int i = 0; i < cnt; i++) {
    printf("%I64d %I64d\n", a[i], b[i]);
  }
  for (int i = cnt - 1; i >= 0; i--) {
    if (b[i] != a[i]) printf("%I64d %I64d\n", b[i], a[i]);
  }
  return 0;
}