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.

<image>

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 mod = 1000000007, oo = 1000000009;
void Emsawy() {}
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, -1, -1, 1};
vector<vector<int> > adj;
const long long N = 1500120;
long long n, m, k, w;
unsigned long long valid(long long i, long long mid) {
  unsigned long long ret = (i * (i + 1) / 2 * (i * 2 + 1)) / 3;
  unsigned long long suf = (i * (i + 1)) / 2;
  if (suf > 2e18 / (mid - i)) return 2e18;
  suf *= (mid - i);
  return ret + suf;
}
int main() {
  Emsawy();
  while (cin >> n) {
    vector<pair<long long, long long> > ans;
    for (int i = 1; i < N; i++) {
      long long s = i, e = n;
      while (s < e) {
        long long mid = s + (e - s + 1) / 2;
        if (valid(i, mid) <= n)
          s = mid;
        else
          e = mid - 1;
      }
      if (valid(i, s) == n) {
        ans.push_back(make_pair(i, s));
      }
    }
    int size = ans.size() * 2;
    if (ans.back().first == ans.back().second) size--;
    cout << size << endl;
    for (int i = 0; i < ans.size(); i++) {
      printf("%lld %lld\n", ans[i].first, ans[i].second);
    }
    for (int i = ans.size() - 1 - (size & 1); i >= 0; i--) {
      cout << ans[i].second << " " << ans[i].first << endl;
    }
  }
  return 0;
}