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;
inline void say(char *msg) { fprintf(stderr, "%s", msg); }
long long n, len, x[10000000], y[10000000];
int main() {
  scanf("%I64d\n", &n);
  for (long long i = 1; i * (i + 1) * (2 * i + 1) / 6 <= n; i++) {
    long long res = n - i * (i + 1) * (2 * i + 1) / 6;
    long long delta = i * (i + 1) / 2;
    if (res % delta == 0) {
      len++;
      x[len] = i;
      y[len] = i + res / delta;
    }
  }
  if (x[len] == y[len]) {
    printf("%I64d\n", 2 * len - 1);
    for (int i = 1; i <= (len); i++) printf("%I64d %I64d\n", x[i], y[i]);
    for (int i = (len - 1); i >= (1); i--) printf("%I64d %I64d\n", y[i], x[i]);
  } else {
    printf("%I64d\n", 2 * len);
    for (int i = 1; i <= (len); i++) printf("%I64d %I64d\n", x[i], y[i]);
    for (int i = (len); i >= (1); i--) printf("%I64d %I64d\n", y[i], x[i]);
  }
  return 0;
}