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;
pair<int, long long> rez[1300000];
int main() {
  long long x;
  scanf("%I64d", &x);
  int k = 0;
  for (int i = 1; 1LL * i * i * i <= 6LL * x; ++i) {
    long long e1 = -1LL * i + 6LL * x + 1LL * i * i * i;
    long long e2 = 3LL * i * (i + 1);
    if (e1 % e2) continue;
    long long n = e1 / e2;
    if (1LL * i > n) break;
    rez[++k] = make_pair(i, n);
  }
  if (1LL * rez[k].first == rez[k].second) {
    printf("%d\n", k * 2 - 1);
    for (int i = 1; i <= k; ++i) {
      printf("%d ", rez[i].first);
      printf("%I64d\n", rez[i].second);
    }
    --k;
  } else {
    printf("%d\n", k * 2);
    for (int i = 1; i <= k; ++i) {
      printf("%d ", rez[i].first);
      printf("%I64d\n", rez[i].second);
    }
  }
  for (int i = k; i; --i) {
    printf("%I64d ", rez[i].second);
    printf("%d\n", rez[i].first);
  }
  return 0;
}