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.

<image>

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;
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
int dxx[] = {+1, 0, -1, 0, +1, +1, -1, -1};
int dyy[] = {0, +1, 0, -1, +1, -1, +1, -1};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
vector<pair<long long, long long> > ans;
int main() {
  long long x, i, sz = 0, l, y;
  scanf("%lld", &x);
  l = min(x, 10000000LL);
  for (i = 1; i <= l; i++) {
    y = x * 6LL;
    if (y % i) continue;
    y /= i;
    if (y % (i + 1)) continue;
    y /= (i + 1);
    y += (i - 1);
    if (y % 3) continue;
    y /= 3;
    if (y < i) continue;
    ans.push_back(make_pair(i, y));
    if (i == y)
      sz++;
    else
      sz += 2;
  }
  printf("%lld\n", sz);
  for (i = 0; i < ans.size(); i++)
    printf("%lld %lld\n", ans[i].first, ans[i].second);
  for (i = ans.size() - 1; ~i; i--)
    if (ans[i].first != ans[i].second)
      printf("%lld %lld\n", ans[i].second, ans[i].first);
  return 0;
}