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>
const double EPS = 1e-10;
using namespace std;
inline int rit() {
  int f = 0, key = 1;
  char c;
  do {
    c = getchar();
    if (c == '-') key = -1;
  } while (c < '0' || c > '9');
  do {
    f = f * 10 + c - '0';
    c = getchar();
  } while (c >= '0' && c <= '9');
  return f * key;
}
vector<pair<long long, long long> > ans;
const int Z = 100005;
long long n;
int calc(long long x) {
  long long f = x * (x + 1) * (x * 2 + 1) / 6, f2 = x * (x + 1) / 2, fff;
  if (f > n) return 0;
  fff = (n - f) / f2;
  if ((n - f) % f2 == 0) ans.push_back(make_pair(x, x + fff));
  return 1;
}
void read() {
  long long i;
  scanf("%lld", &n);
  for (i = 1; i <= 3000000; i++) {
    if (!calc(i)) break;
  }
  sort(ans.begin(), ans.end());
  if (ans.size() == 0) {
    cout << 0 << endl;
  } else {
    printf(
        "%lld\n",
        ans.size() * 2 -
            (ans[ans.size() - 1].first == ans[ans.size() - 1].second ? 1 : 0));
    for (i = 0; i < ans.size(); i++) {
      printf("%lld %lld\n", ans[i].first, ans[i].second);
    }
    if (ans[i - 1].first == ans[i - 1].second) i--;
    for (i--; i >= 0; i--) {
      printf("%lld %lld\n", ans[i].second, ans[i].first);
    }
  }
}
int main() {
  read();
  return 0;
}