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;
struct Ans {
  long long a, b;
} ans[2000010];
int sum = 0;
long long x;
int xx;
int main() {
  scanf("%lld", &x);
  x *= 6;
  int n = pow((double)x, (double)1 / 3) + 1;
  for (long long i = 1; i <= n; i++) {
    long long y = (long long)i * (i + 1);
    if (!(x % y)) {
      long long j = (long long)x / y;
      if (!((j + i - 1) % 3)) {
        long long k = (long long)(j + i - 1) / 3;
        if (i <= k) {
          ans[++sum].a = i;
          ans[sum].b = k;
        }
      }
    }
  }
  if (ans[sum].a != ans[sum].b)
    printf("%d\n", sum * 2);
  else
    printf("%d\n", sum * 2 - 1);
  for (int i = 1; i <= sum; i++) printf("%lld %lld\n", ans[i].a, ans[i].b);
  if (ans[sum].a != ans[sum].b) printf("%lld %lld\n", ans[sum].b, ans[sum].a);
  for (int i = sum - 1; i >= 1; i--) printf("%lld %lld\n", ans[i].b, ans[i].a);
  return 0;
}