#include <bits/stdc++.h>
using namespace std;
int main() {
  int m;
  scanf("%d", &m);
  int cnt2 = 0, cnt5 = 0;
  int l = -1, r;
  for (int i = 1;; ++i) {
    int t = i;
    while (t % 2 == 0) cnt2++, t /= 2;
    while (t % 5 == 0) cnt5++, t /= 5;
    if (l == -1 && min(cnt2, cnt5) >= m) l = i;
    if (min(cnt2, cnt5) > m) {
      r = i - 1;
      break;
    }
  }
  printf("%d\n", r - l + 1);
  for (int i = l; i <= r; ++i) printf("%d ", i);
  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;
int main() {
  int m;
  scanf("%d", &m);
  int cnt2 = 0, cnt5 = 0;
  int l = -1, r;
  for (int i = 1;; ++i) {
    int t = i;
    while (t % 2 == 0) cnt2++, t /= 2;
    while (t % 5 == 0) cnt5++, t /= 5;
    if (l == -1 && min(cnt2, cnt5) >= m) l = i;
    if (min(cnt2, cnt5) > m) {
      r = i - 1;
      break;
    }
  }
  printf("%d\n", r - l + 1);
  for (int i = l; i <= r; ++i) printf("%d ", i);
  return 0;
}

```