#include <bits/stdc++.h>
const double EPS = 0.00000001;
const long long mod = 1000000000 + 7;
using namespace std;
int f(long long n) {
  if (n < 5) return 0;
  return f(n / 5) + (n / 5);
}
int main() {
  int n, b = 0;
  cin >> n;
  long long x = 0;
  for (long long i = 5;; i += 5) {
    x = f(i);
    if (x == n) {
      cout << 5 << endl;
      for (int j = i; j < i + 5; j++) cout << j << " ";
      b = 1;
      break;
    } else if (x > n)
      break;
  }
  if (!b) cout << 0 << endl;
  int ____________;
  cin >> ____________;
  return 0;
}

### Prompt

Create a solution in CPP for 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>
const double EPS = 0.00000001;
const long long mod = 1000000000 + 7;
using namespace std;
int f(long long n) {
  if (n < 5) return 0;
  return f(n / 5) + (n / 5);
}
int main() {
  int n, b = 0;
  cin >> n;
  long long x = 0;
  for (long long i = 5;; i += 5) {
    x = f(i);
    if (x == n) {
      cout << 5 << endl;
      for (int j = i; j < i + 5; j++) cout << j << " ";
      b = 1;
      break;
    } else if (x > n)
      break;
  }
  if (!b) cout << 0 << endl;
  int ____________;
  cin >> ____________;
  return 0;
}

```