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;
struct dat {
  long long x, y;
  dat(long long x = 0, long long y = 0) : x(x), y(y) {}
} p[1000000];
bool operator<(dat A, dat B) {
  if (A.x != B.x) return A.x < B.x;
  return A.y < B.y;
}
bool operator==(dat A, dat B) {
  if (A.x == B.x && A.y == B.y) return 1;
  return 0;
}
int main() {
  long long x;
  scanf("%I64d\n", &x);
  int cnt = 0;
  for (long long i = 1; i <= 2e6; i++) {
    if (6 * x % (i * (i + 1)) == 0) {
      long long y = 6 * x / i / (i + 1);
      if ((2 * i - 2) > y) continue;
      if ((y - 2 * i + 2) % 3) continue;
      long long z = (y - 2 * i + 2) / 3;
      long long j = i + z - 1;
      p[cnt++] = dat(i, j);
      p[cnt++] = dat(j, i);
    }
  }
  sort(p, p + cnt);
  cnt = unique(p, p + cnt) - p;
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) {
    printf("%I64d %I64d\n", p[i].x, p[i].y);
  }
}