#include <bits/stdc++.h>
using namespace std;
int f(int n) {
  int ans = 0;
  for (int i = 5; n / i >= 1; i *= 5) ans += n / i;
  return ans;
}
int n, l, r = 200000, ans, m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  while (l <= r) {
    m = (l + r) / 2;
    ans = f(m * 5);
    if (ans == n) {
      ans = m * 5;
      break;
    } else if (ans < n)
      l = m + 1;
    else
      r = m - 1;
  }
  if (l <= r)
    if (n != 0)
      cout << 5 << endl
           << ans << " " << ans + 1 << " " << ans + 2 << " " << ans + 3 << " "
           << ans + 4;
    else
      cout << 4 << endl
           << ans + 1 << " " << ans + 2 << " " << ans + 3 << " " << ans + 4;
  else
    cout << 0;
  return 0;
}

### Prompt

Your challenge is to write 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 f(int n) {
  int ans = 0;
  for (int i = 5; n / i >= 1; i *= 5) ans += n / i;
  return ans;
}
int n, l, r = 200000, ans, m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  while (l <= r) {
    m = (l + r) / 2;
    ans = f(m * 5);
    if (ans == n) {
      ans = m * 5;
      break;
    } else if (ans < n)
      l = m + 1;
    else
      r = m - 1;
  }
  if (l <= r)
    if (n != 0)
      cout << 5 << endl
           << ans << " " << ans + 1 << " " << ans + 2 << " " << ans + 3 << " "
           << ans + 4;
    else
      cout << 4 << endl
           << ans + 1 << " " << ans + 2 << " " << ans + 3 << " " << ans + 4;
  else
    cout << 0;
  return 0;
}

```