#include <bits/stdc++.h>
using namespace std;
int main() {
  int m;
  cin >> m;
  vector<int> r;
  int t = 0;
  int f = 0;
  for (int i = 1; i <= 10 * m; i++) {
    int x = i;
    while (x % 2 == 0) {
      t++;
      x /= 2;
    }
    while (x % 5 == 0) {
      f++;
      x /= 5;
    }
    if (min(t, f) == m) {
      r.push_back(i);
    }
  }
  printf("%d\n", (int)r.size());
  for (int i = 0; i < r.size(); i++) {
    printf(" %d", r[i]);
  }
  printf("\n");
}

### 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 main() {
  int m;
  cin >> m;
  vector<int> r;
  int t = 0;
  int f = 0;
  for (int i = 1; i <= 10 * m; i++) {
    int x = i;
    while (x % 2 == 0) {
      t++;
      x /= 2;
    }
    while (x % 5 == 0) {
      f++;
      x /= 5;
    }
    if (min(t, f) == m) {
      r.push_back(i);
    }
  }
  printf("%d\n", (int)r.size());
  for (int i = 0; i < r.size(); i++) {
    printf(" %d", r[i]);
  }
  printf("\n");
}

```