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>
void pre();
void post();
using namespace std;
const int MAXN = 1234567;
const int INF = 1000000123;
long long x;
void read() { cin >> x; }
long long summa[1442252];
int main() {
  pre();
  read();
  long long speed = 1;
  vector<pair<long long, long long> > ans;
  ans.reserve(1442252);
  for (long long i = 1; i < 1442252; i++) {
    summa[i] = summa[i - 1] + i * i;
  }
  for (long long i = 1; i + (x - summa[i]) / speed >= i; i++) {
    speed = (i) * (i + 1) / 2;
    if ((x - summa[i]) % speed == 0) {
      ans.push_back(make_pair(i, i + (x - summa[i]) / speed));
    }
  }
  int sizes2 = ans.size();
  for (int i = 0; i < sizes2; i++) {
    ans.push_back(make_pair(ans[i].second, ans[i].first));
  }
  sort(ans.begin(), ans.end());
  ans.erase(unique(ans.begin(), ans.end()), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  }
  post();
  return 0;
}
void pre() { return; }
void post() {}