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 del = 10723;
const long long mod = 1e9 + 7;
const long long maxn = 1e7 + 500;
vector<pair<long long, long long> > ans;
long long g[maxn];
long long c[maxn];
int main() {
  g[1] = 1;
  g[2] = 5;
  for (long long i = 3; i < 1e7 + 6; i++) {
    g[i] = g[i - 1] + (i * i);
  }
  for (long long i = 1; i < 1e7 + 6; i++) {
    c[i] = i * (i + 1) / 2;
  }
  long long n = 8e5;
  long long x;
  scanf("%I64d", &x);
  for (long long i = 1; i < 1e7; i++) {
    if ((x - g[i]) % c[i] == 0) {
      long long b = (x - g[i]) / c[i] + i;
      if (i <= b) {
        if (i != b) {
          ans.push_back(make_pair(i, b));
          ans.push_back(make_pair(b, i));
        } else {
          ans.push_back(make_pair(i, b));
        }
      }
    }
  }
  printf("%i\n", ans.size());
  sort(ans.begin(), ans.end());
  for (long long i = 0; i < ans.size(); i++) {
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  }
}