#include <bits/stdc++.h>
using namespace std;
int sp[50007];
int func(int x) {
  int i = 5, res = 0;
  while (x / i > 0) {
    res = res + (x / i);
    i = i * 5;
  }
  return res;
}
int main() {
  int k, l, r, m, i;
  scanf("%d", &k);
  l = 0;
  r = 1000000;
  while (l < r) {
    m = l + (r - l) / 2;
    if (func(m) < k)
      l = m + 1;
    else
      r = m;
  }
  if (func(l) == k) {
    printf("5\n");
    for (i = 1; i <= 5; i++) printf("%d ", l++);
  } else
    printf("0\n");
  return 0;
}

### Prompt

Your task is to create 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 sp[50007];
int func(int x) {
  int i = 5, res = 0;
  while (x / i > 0) {
    res = res + (x / i);
    i = i * 5;
  }
  return res;
}
int main() {
  int k, l, r, m, i;
  scanf("%d", &k);
  l = 0;
  r = 1000000;
  while (l < r) {
    m = l + (r - l) / 2;
    if (func(m) < k)
      l = m + 1;
    else
      r = m;
  }
  if (func(l) == k) {
    printf("5\n");
    for (i = 1; i <= 5; i++) printf("%d ", l++);
  } else
    printf("0\n");
  return 0;
}

```