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 gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = (res * a) % ((int)(1e9) + 7);
    a = (a * a) % ((int)(1e9) + 7);
    b /= 2;
  }
  return res;
}
int tree[123456];
void update(int idx, int val) {
  while (idx < 123456) {
    tree[idx] += val;
    idx += idx & -idx;
  }
}
int qry(int idx) {
  int sum = 0;
  while (idx) {
    sum += tree[idx];
    idx -= idx & -idx;
  }
  return sum;
}
int main() {
  long long x;
  ios::sync_with_stdio(false);
  cin >> x;
  set<pair<long long, long long> > arr;
  long long newnumber;
  long long n = 1;
  for (;;) {
    newnumber = (n * (n + 1) * (2 * n + 1)) / 6;
    if (newnumber > x) break;
    long long curnumber = x - newnumber;
    if (curnumber % (n * (n + 1) / 2) == 0) {
      long long y = n;
      long long z = n + curnumber / (n * (n + 1) / 2);
      arr.insert(make_pair(y, z));
      arr.insert(make_pair(z, y));
    }
    n++;
  }
  vector<pair<long long, long long> > ans;
  for (set<pair<long long, long long> >::iterator it = arr.begin();
       it != arr.end(); it++)
    ans.push_back(make_pair(it->first, it->second));
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << " " << ans[i].second << "\n";
  return 0;
}