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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MXN = 3e6 + 10;
long long m, r, c, x, MX = 3e6, A[MXN];
long double P[MXN];
long long Solve(long long n) {
  long long t = (long long)(x - P[n - 1] +
                            (((long double)n) * n * (n - 1) / 2)),
            k = (n * n + n) / 2;
  if (t % k != 0 || t < 0) return (-1);
  return (t / k);
}
int main() {
  scanf("%I64d", &x);
  for (int i = 1; i <= MX; i++) {
    P[i] = P[i - 1] + (1ll * i) * (1ll * i);
  }
  for (int i = 2; i <= MX; i++) {
    r = Solve(1ll * i);
    if (r < i) continue;
    if (r == -1) continue;
    A[i] = r;
    if (i != r) c++;
    c++;
  }
  A[1] = x;
  if (x != 1) c++;
  c++;
  printf("%I64d\n", c);
  for (int i = 1; i <= MX; i++)
    if (A[i] != 0) printf("%d %I64d\n", i, A[i]);
  for (int i = MX; i >= 1; i--)
    if (A[i] != 0 && A[i] != i) printf("%I64d %d\n", A[i], i);
  return (0);
}