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 a[2000005], b[2000005];
long long ans1[2000005], ans2[2000005];
int main() {
  a[0] = 0;
  b[0] = 0;
  for (int i = 0; i < 2000002; i++) {
    a[i] = a[i - 1] + i;
    b[i] = b[i - 1] + (long long)i * (i - 1);
  }
  long long x;
  while (scanf("%I64d\n", &x) != EOF) {
    int cnt = 0, rp = 0;
    ;
    for (int i = 1; i < 2000002; i++) {
      if (b[i] >= x) break;
      if ((x - b[i]) % a[i] == 0) {
        ans1[cnt] = i;
        ans2[cnt] = (x - b[i]) / a[i] + i - 1;
        if (ans1[cnt] == ans2[cnt]) rp = 1;
        cnt++;
      }
    }
    printf("%d\n", cnt * 2 - rp);
    for (int i = 0; i < cnt; i++) printf("%I64d %I64d\n", ans1[i], ans2[i]);
    for (int i = cnt - 1 - rp; i >= 0; i--)
      printf("%I64d %I64d\n", ans2[i], ans1[i]);
  }
  return 0;
}