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;
long long n;
vector<pair<long long, long long> > v;
bool same = false;
int main() {
  scanf("%I64d", &n);
  long long counter = 0;
  long long triangle = 0;
  for (long long i = 1; i <= sqrt(n); i++) {
    counter += i * i;
    triangle += i;
    if (n - counter < 0) break;
    if ((n - counter) % (triangle) == 0) {
      v.push_back(make_pair(i, (n - counter) / (triangle) + i));
      if (i == (n - counter) / triangle + i) {
        same = true;
      }
    }
  }
  if (same == true) {
    printf("%d\n", (int)2 * v.size() - 1);
    for (vector<pair<long long, long long> >::iterator it = v.begin();
         it != v.end(); it++) {
      printf("%I64d %I64d\n", it->first, it->second);
    }
    for (vector<pair<long long, long long> >::reverse_iterator it = v.rbegin();
         it != v.rend(); it++) {
      if (it == v.rbegin()) continue;
      printf("%I64d %I64d\n", it->second, it->first);
    }
    return 0;
  }
  printf("%d\n", (int)2 * v.size());
  for (vector<pair<long long, long long> >::iterator it = v.begin();
       it != v.end(); it++) {
    printf("%I64d %I64d\n", it->first, it->second);
  }
  for (vector<pair<long long, long long> >::reverse_iterator it = v.rbegin();
       it != v.rend(); it++) {
    printf("%I64d %I64d\n", it->second, it->first);
  }
}