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;
long long x, squ, res, cnt;
pair<long long, long long> ans[1000005];
int main() {
  scanf("%I64d", &x);
  for (long long n = 1; squ = n * (n + 1) * (2 * n + 1) / 6, squ <= x; n++) {
    long long res = n * (n + 1) / 2;
    if ((x - squ) % res == 0) ans[++cnt] = make_pair(n, n + (x - squ) / res);
  }
  bool flag = false;
  if (ans[cnt].first == ans[cnt].second) flag = true;
  printf("%I64d\n", flag ? cnt * 2 - 1 : cnt * 2);
  for (long long i = 1; i <= cnt; i++)
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  for (long long i = cnt; i >= 1; i--) {
    if (flag && i == cnt)
      continue;
    else
      printf("%I64d %I64d\n", ans[i].second, ans[i].first);
  }
  return 0;
}