#include <bits/stdc++.h>
using namespace std;
long long n, m, d, a, b, c, num1;
const long long mod = 1000 * 1000 * 1000 + 7;
vector<int> num;
int main() {
  scanf("%d", &m);
  for (int i = 1;; i++) {
    if (!(i % 5) || !(i % 10)) {
      int k = i;
      if (!(i % 5))
        while (!(k % 5)) k /= 5, num1++;
    }
    if (num1 == m) num.push_back(i);
    if (num1 > m) break;
  }
  printf("%d\n", num.size());
  for (int i = 0; i < num.size(); i++) printf("%d ", num[i]);
}

### Prompt

Your task is to create 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 n, m, d, a, b, c, num1;
const long long mod = 1000 * 1000 * 1000 + 7;
vector<int> num;
int main() {
  scanf("%d", &m);
  for (int i = 1;; i++) {
    if (!(i % 5) || !(i % 10)) {
      int k = i;
      if (!(i % 5))
        while (!(k % 5)) k /= 5, num1++;
    }
    if (num1 == m) num.push_back(i);
    if (num1 > m) break;
  }
  printf("%d\n", num.size());
  for (int i = 0; i < num.size(); i++) printf("%d ", num[i]);
}

```