#include <bits/stdc++.h>
using namespace std;
int n, m, two, five, x, ans[1000005], i;
int main() {
  scanf("%d", &m);
  for (n = 1;; n++) {
    for (x = n; x % 2 == 0; x /= 2) two++;
    for (x = n; x % 5 == 0; x /= 5) five++;
    if (min(two, five) > m) break;
    if (min(two, five) == m) ans[++*ans] = n;
  }
  printf("%d\n", *ans);
  for (i = 1; i <= *ans; i++) printf("%d ", ans[i]);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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, m, two, five, x, ans[1000005], i;
int main() {
  scanf("%d", &m);
  for (n = 1;; n++) {
    for (x = n; x % 2 == 0; x /= 2) two++;
    for (x = n; x % 5 == 0; x /= 5) five++;
    if (min(two, five) > m) break;
    if (min(two, five) == m) ans[++*ans] = n;
  }
  printf("%d\n", *ans);
  for (i = 1; i <= *ans; i++) printf("%d ", ans[i]);
}

```