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 maxn = 1e5 + 5;
const int maxlog = 22;
const long long inf = 1e18;
const int mod = 1e9 + 7;
const int sz = 350;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long x;
  cin >> x;
  long long a, b, c, d, e;
  vector<pair<long long, long long> > v;
  vector<pair<long long, long long> > g;
  for (long long i = 1; i <= 6000006; i++) {
    a = i * (i - 1) / 2;
    b = i * (i - 1) * (2 * i - 1) / 6;
    c = x;
    c += (a * i);
    c -= (b);
    d = (i * i) - (a);
    e = c / d;
    if (i > e) break;
    if (c % d == 0) v.push_back({i, e});
    if (i != e && c % d == 0) g.push_back({e, i});
  }
  cout << v.size() + g.size() << endl;
  for (int i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << endl;
  for (int i = g.size() - 1; i >= 0; i--)
    cout << g[i].first << " " << g[i].second << endl;
  return 0;
}