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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const long long LINF = (1LL << 60);
const int INF = 1 << 30;
const int NS = 2000000;
const int MS = 19;
const int MOD = 1000000007;
const double PI = acos(-1.0);
int main() {
  long long n;
  vector<pair<long long, long long> > ans;
  while (cin >> n) {
    ans.clear();
    long long m = n * 6;
    for (long long i = 1; i < NS; i++) {
      long long d = i * (i + 1);
      long long t = d * (i + i);
      if (t > m) {
        break;
      }
      if (0 == (m % d)) {
        long long w = i;
        long long k = m / d;
        long long h = (k - 1 + w) / 3;
        if (m == (d * (h * 3 - w + 1))) {
          ans.push_back(make_pair(w, h));
          if (w != h) {
            ans.push_back(make_pair(h, w));
          }
        }
      }
    }
    sort(ans.begin(), ans.end());
    int num = ans.size();
    cout << num << endl;
    for (int i = 0; i < num; i++) {
      cout << ans[i].first << " " << ans[i].second << endl;
    }
  }
  return 0;
}