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;
const long long inf = 1000000000ll * 1000000000ll;
const long long nmax = 1442251;
int main() {
  long long x;
  scanf("%I64d", &x);
  vector<long long> sumSqr(nmax);
  for (long long i = 1; i < nmax; ++i) {
    sumSqr[i] = sumSqr[i - 1] + i * i;
  }
  vector<pair<long long, long long> > ans;
  for (long long n = 1; n < nmax; ++n) {
    long long r = x - sumSqr[n - 1];
    long long t = n * (n - 1) / 2;
    r += t * n;
    if (r < 0) {
      continue;
    }
    r *= 2;
    if (r % (n * (n + 1))) {
      continue;
    }
    long long m = (r / n) / (n + 1);
    if (m < n) {
      continue;
    }
    ans.push_back(make_pair(n, m));
    if (n != m) {
      ans.push_back(make_pair(m, n));
    }
  }
  sort((ans).begin(), (ans).end());
  printf("%d\n", (int)(ans).size());
  for (int i = 0; i < (int)(ans).size(); ++i) {
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  }
}