#include <bits/stdc++.h>
const int N = 1e7 + 100;
int m;
int f[N];
void init() { std::cin >> m; }
void work() {
  f[0] = 0;
  for (int i = 1; i <= 10000000; i++) {
    f[i] = i / 5 + f[i / 5];
  }
  int count = 0;
  for (int i = 1; i <= 10000000; i++) {
    if (f[i] == m) {
      count++;
    }
  }
  std::cout << count << std::endl;
  for (int i = 1; i <= 10000000; i++) {
    if (f[i] == m) {
      printf("%d ", i);
    }
  }
  printf("\n");
}
int main() {
  init();
  work();
  return 0;
}

### 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>
const int N = 1e7 + 100;
int m;
int f[N];
void init() { std::cin >> m; }
void work() {
  f[0] = 0;
  for (int i = 1; i <= 10000000; i++) {
    f[i] = i / 5 + f[i / 5];
  }
  int count = 0;
  for (int i = 1; i <= 10000000; i++) {
    if (f[i] == m) {
      count++;
    }
  }
  std::cout << count << std::endl;
  for (int i = 1; i <= 10000000; i++) {
    if (f[i] == m) {
      printf("%d ", i);
    }
  }
  printf("\n");
}
int main() {
  init();
  work();
  return 0;
}

```