#include <bits/stdc++.h>
using namespace std;
long long nr5, m;
vector<long long> sol;
int main() {
  long long nr;
  cin >> m;
  for (nr = 1; nr5 < m; nr++) {
    long long aux = nr;
    while (aux % 5 == 0) {
      aux /= 5;
      nr5++;
    }
    if (nr5 == m) break;
  }
  if (nr5 > m)
    cout << 0;
  else {
    for (; nr5 == m;) {
      sol.push_back(nr);
      nr++;
      if (nr % 5 == 0) nr5++;
    }
    cout << sol.size() << "\n";
    for (auto x : sol) cout << x << ' ';
  }
  return 0;
}

### Prompt

Please formulate a CPP solution 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;
long long nr5, m;
vector<long long> sol;
int main() {
  long long nr;
  cin >> m;
  for (nr = 1; nr5 < m; nr++) {
    long long aux = nr;
    while (aux % 5 == 0) {
      aux /= 5;
      nr5++;
    }
    if (nr5 == m) break;
  }
  if (nr5 > m)
    cout << 0;
  else {
    for (; nr5 == m;) {
      sol.push_back(nr);
      nr++;
      if (nr % 5 == 0) nr5++;
    }
    cout << sol.size() << "\n";
    for (auto x : sol) cout << x << ' ';
  }
  return 0;
}

```