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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = -1;
const long long INF = 1e9 + 19;
void read() {
  unsigned long long x;
  cin >> x;
  vector<pair<unsigned long long, unsigned long long> > answer;
  unsigned long long sum = 0;
  for (unsigned long long b = 1;; b++) {
    sum += (b - 1) * 1ll * (b - 1);
    unsigned long long tmp = x - sum;
    tmp += b * 1ll * (b - 1) * 1ll * b / 2;
    if (tmp <= 0) break;
    unsigned long long d = b * 1ll * b - b * 1ll * (b - 1) / 2;
    if (tmp / d < b) break;
    if (tmp % d == 0) {
      answer.push_back(make_pair(b, tmp / d));
      if (b != tmp / d) answer.push_back(make_pair(tmp / d, b));
    }
  }
  sort(answer.begin(), answer.end());
  printf("%d\n", (int)answer.size());
  for (auto x : answer) printf("%lld %lld\n", x.first, x.second);
}
void solve() {}
void printAns() {}
void stress() {}
int main() {
  if (1) {
    int k = 1;
    for (int tt = 0; tt < k; tt++) {
      read();
      solve();
      printAns();
    }
  } else {
    stress();
  }
  return 0;
}