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;
const int N = 2e6 + 10;
long long x, in, cnt, ans;
long long sum[N];
signed main() {
  for (long long i = 1; i <= 2e6; i++) sum[i] = sum[i - 1] + i * i;
  cin >> x;
  for (long long i = 1; i <= 2e6; i++) {
    in = sum[i];
    if (in > x) break;
    if (in == x)
      ans++;
    else {
      if ((x - in) % ((i * (i + 1)) / 2) != 0) continue;
      cnt = (x - in) / ((i * (i + 1)) / 2);
      ans += 2;
    }
  }
  cout << ans << '\n';
  for (long long i = 1; i <= 2e6; i++) {
    in = sum[i];
    if (in > x) break;
    if (in == x)
      cout << i << " " << i << '\n';
    else {
      if ((x - in) % ((i * (i + 1)) / 2) != 0) continue;
      cnt = (x - in) / ((i * (i + 1)) / 2);
      cout << i << " " << i + cnt << '\n';
    }
  }
  for (long long i = 2e6; i >= 1; i--) {
    in = sum[i];
    if (in > x)
      continue;
    else if (in != x) {
      if ((x - in) % ((i * (i + 1)) / 2) != 0) continue;
      cnt = (x - in) / ((i * (i + 1)) / 2);
      cout << i + cnt << " " << i << '\n';
    }
  }
  return 0;
}