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 node {
  long long x, y;
  friend bool operator<(const node &a, const node &b) {
    return a.x == b.x ? a.y < b.y : a.x < b.x;
  }
} a[1500010];
long long k;
long long t;
int main() {
  cin >> k;
  long long l = 1, r = 1500000;
  while (l + 1 < r) {
    long long mid = (l + r) / 2;
    if (2 * mid * mid * mid + 3 * mid * mid + mid <= 6 * k)
      l = mid;
    else
      r = mid - 1;
  }
  if (2 * r * r * r + 3 * r * r + r <= 6 * k)
    t = r;
  else
    t = l;
  int ret = 0;
  for (long long i = 1; i <= t; ++i)
    if (6 * k % i == 0) {
      long long j = (6 * k + i * i * i - i) / i / 3 / (i + 1);
      long long cnt = i * (3 * i * j - i * i + 3 * j + 1);
      if (cnt == 6 * k) {
        if (i == j)
          a[++ret].x = i, a[ret].y = j;
        else
          a[++ret].x = i, a[ret].y = j, a[++ret].x = j, a[ret].y = i;
      }
    }
  sort(a + 1, a + ret + 1);
  printf("%d\n", ret);
  for (int i(1); i <= (ret); ++i) printf("%lld %lld\n", a[i].x, a[i].y);
  return 0;
}