#include <bits/stdc++.h>
using namespace std;
int s(int n) {
  int sum = 0;
  for (int i = 5; i <= n; i *= 5) sum += n / i;
  return sum;
}
int main() {
  int n;
  scanf("%d", &n);
  int lo = 0, hi = 1e6 + 5, mid;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    int sum = s(mid);
    if (sum > n)
      hi = mid - 1;
    else if (sum < n)
      lo = mid + 1;
    else
      break;
  }
  if (s(mid) != n) {
    printf("0");
    return 0;
  }
  printf("5\n");
  for (int i = mid - (mid % 5); i < mid - (mid % 5) + 5; i++) printf("%d ", i);
}

### Prompt

Generate 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;
int s(int n) {
  int sum = 0;
  for (int i = 5; i <= n; i *= 5) sum += n / i;
  return sum;
}
int main() {
  int n;
  scanf("%d", &n);
  int lo = 0, hi = 1e6 + 5, mid;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    int sum = s(mid);
    if (sum > n)
      hi = mid - 1;
    else if (sum < n)
      lo = mid + 1;
    else
      break;
  }
  if (s(mid) != n) {
    printf("0");
    return 0;
  }
  printf("5\n");
  for (int i = mid - (mid % 5); i < mid - (mid % 5) + 5; i++) printf("%d ", i);
}

```