#include <bits/stdc++.h>
int trailing_zeros(int a) {
  int i;
  int counter = 0;
  for (i = 5; i <= a; i *= 5) {
    counter = counter + (a / i);
  }
  return counter;
}
int storage[1000001];
int main() {
  int input;
  scanf("%d", &input);
  int cnt = 0;
  int a, b, c, d, e;
  b = 0;
  for (a = 1; a <= 1000001; a++) {
    int x = trailing_zeros(a);
    if (x == input) {
      storage[b++] = a;
      cnt++;
    }
  }
  printf("%d\n", cnt);
  for (a = 0; a < b; a++) {
    printf("%d ", storage[a]);
  }
}

### 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>
int trailing_zeros(int a) {
  int i;
  int counter = 0;
  for (i = 5; i <= a; i *= 5) {
    counter = counter + (a / i);
  }
  return counter;
}
int storage[1000001];
int main() {
  int input;
  scanf("%d", &input);
  int cnt = 0;
  int a, b, c, d, e;
  b = 0;
  for (a = 1; a <= 1000001; a++) {
    int x = trailing_zeros(a);
    if (x == input) {
      storage[b++] = a;
      cnt++;
    }
  }
  printf("%d\n", cnt);
  for (a = 0; a < b; a++) {
    printf("%d ", storage[a]);
  }
}

```