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;
long long ans;
struct node {
  long long x, y;
  bool operator<(const node &z) const {
    if (x == z.x) return y < z.y;
    return x < z.x;
  }
  bool operator==(const node &z) const { return x == z.x && y == z.y; }
} a[500010];
int main() {
  long long i, x, y;
  scanf("%I64d", &x);
  x *= 6;
  for (i = 1; i * i * i <= x; ++i) {
    if (x % (i * (i + 1))) continue;
    y = x / (i * (i + 1)) + i - 1;
    if (y / 3 < i) continue;
    if (!(y % 3)) {
      ++ans, a[ans].x = i, a[ans].y = y / 3;
      ++ans, a[ans].x = y / 3, a[ans].y = i;
    }
  }
  sort(a + 1, a + 1 + ans);
  ans = unique(a + 1, a + 1 + ans) - (a + 1);
  printf("%I64d\n", ans);
  for (i = 1; i <= ans; ++i) {
    printf("%I64d %I64d\n", a[i].x, a[i].y);
  }
  return 0;
}