#include <bits/stdc++.h>
using namespace std;
int n;
int solve(int x) {
  int c = 0;
  for (int i = 5; x / i >= 1; i *= 5) c += x / i;
  return c;
}
int main() {
  cin >> n;
  int f = 0, l = 1e9, mid;
  bool b = false;
  while (f < l) {
    mid = (f + l) / 2;
    int temp = solve(mid);
    if (temp < n)
      f = mid + 1;
    else
      l = mid;
  }
  if (solve(l) == n)
    b = true, mid = l;
  else if (solve(mid) == n)
    b = true;
  if (!b) {
    cout << 0 << endl;
    return 0;
  }
  cout << 5 << endl;
  for (int i = 0; i < 5; i++) cout << mid + i << " ";
  cout << endl;
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?

Input

The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.

Output

First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.

Examples

Input

1


Output

5
5 6 7 8 9 

Input

5


Output

0

Note

The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.

In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
int solve(int x) {
  int c = 0;
  for (int i = 5; x / i >= 1; i *= 5) c += x / i;
  return c;
}
int main() {
  cin >> n;
  int f = 0, l = 1e9, mid;
  bool b = false;
  while (f < l) {
    mid = (f + l) / 2;
    int temp = solve(mid);
    if (temp < n)
      f = mid + 1;
    else
      l = mid;
  }
  if (solve(l) == n)
    b = true, mid = l;
  else if (solve(mid) == n)
    b = true;
  if (!b) {
    cout << 0 << endl;
    return 0;
  }
  cout << 5 << endl;
  for (int i = 0; i < 5; i++) cout << mid + i << " ";
  cout << endl;
  return 0;
}

```