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 long long mod = 1000000007;
const int inf = 100000000;
set<pair<long long, long long> > myset;
int main() {
  long long x;
  scanf("%I64d", &x);
  long long upto = 3000000;
  long long i;
  for (i = 1; i <= upto; i++) {
    long long upper = 6 * x - i;
    upper = upper + i * i * i;
    long long lower = 3 * i * i + 3 * i;
    if ((upper % lower) == 0) {
      long long ans = upper / lower;
      if (ans < i) continue;
      if (ans != i) {
        myset.insert(pair<long long, long long>(i, ans));
        myset.insert(pair<long long, long long>(ans, i));
      } else
        myset.insert(pair<long long, long long>(i, ans));
    }
  }
  set<pair<long long, long long> >::iterator it;
  long long sz = myset.size();
  printf("%I64d\n", sz);
  for (it = myset.begin(); it != myset.end(); it++) {
    printf("%I64d %I64d\n", it->first, it->second);
  }
  return 0;
}