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;
const int INF = 1e9;
const int mod = 1e5 + 3;
const long long int LONGINF = 4e18;
const double eps = 1e-9;
const double PI = 3.1415926535897932384626433832795;
void solve() {
  long long int a;
  scanf("%I64d", &a);
  vector<pair<long long int, long long int>> v;
  for (long long int n = 1;
       n * (n - 1) * (2 * n - 1) / 6 - n * n * (n - 1) / 2 <= a; n++) {
    long long int m = a - (n * (n - 1) * (2 * n - 1) / 6 - n * n * (n - 1) / 2);
    if (m % (n * (n + 1) / 2) == 0 && m / (n * (n + 1) / 2) >= n) {
      v.push_back(make_pair(n, m / (n * (n + 1) / 2)));
    }
  }
  printf("%d\n", 2 * v.size() - (int)(v.back().first == v.back().second));
  for (int i = 0; i < (v.size()); i++) {
    printf("%I64d %I64d\n", v[i].first, v[i].second);
  }
  for (int i = v.size() - 1 - (int)(v.back().first == v.back().second); i >= 0;
       i--) {
    printf("%I64d %I64d\n", v[i].second, v[i].first);
  }
}
int main() {
#pragma comment(linker, "/STACK:268435456")
  solve();
  return 0;
}