#include <bits/stdc++.h>
using namespace std;
long long maxi = LLONG_MIN;
long long mini = LLONG_MAX;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int main() {
  int m;
  cin >> m;
  int f = 0;
  long long ans = -1;
  long long n = 0;
  long long c = 0;
  while (c <= m) {
    c = 0;
    for (long long i = 5; n / i >= 1; i *= 5) {
      c += n / i;
    }
    if (c == m) {
      ans = n;
      f = 1;
      break;
    } else {
      ++n;
    }
  }
  if (f == 0) {
    cout << "0";
  } else {
    cout << "5" << '\n';
    cout << ans << " " << ans + 1 << " " << ans + 2 << " " << ans + 3 << " "
         << ans + 4;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long maxi = LLONG_MIN;
long long mini = LLONG_MAX;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int main() {
  int m;
  cin >> m;
  int f = 0;
  long long ans = -1;
  long long n = 0;
  long long c = 0;
  while (c <= m) {
    c = 0;
    for (long long i = 5; n / i >= 1; i *= 5) {
      c += n / i;
    }
    if (c == m) {
      ans = n;
      f = 1;
      break;
    } else {
      ++n;
    }
  }
  if (f == 0) {
    cout << "0";
  } else {
    cout << "5" << '\n';
    cout << ans << " " << ans + 1 << " " << ans + 2 << " " << ans + 3 << " "
         << ans + 4;
  }
  return 0;
}

```