#include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) { return (y == 0) ? x : gcd(y, x % y); }
long long lcm(long long x, long long y) { return x * (y / gcd(x, y)); }
int findTrailingZeros(int n) {
  int count = 0;
  for (int i = 5; n / i >= 1; i *= 5) count += n / i;
  return count;
}
int main() {
  int m;
  cin >> m;
  set<int> ans;
  for (int i = 0; i <= 1000000; i++)
    if (findTrailingZeros(i) == m) ans.insert(i);
  cout << ans.size() << endl;
  for (set<int>::iterator t = ans.begin(); t != ans.end(); t++)
    cout << *t << " ";
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 gcd(long long x, long long y) { return (y == 0) ? x : gcd(y, x % y); }
long long lcm(long long x, long long y) { return x * (y / gcd(x, y)); }
int findTrailingZeros(int n) {
  int count = 0;
  for (int i = 5; n / i >= 1; i *= 5) count += n / i;
  return count;
}
int main() {
  int m;
  cin >> m;
  set<int> ans;
  for (int i = 0; i <= 1000000; i++)
    if (findTrailingZeros(i) == m) ans.insert(i);
  cout << ans.size() << endl;
  for (set<int>::iterator t = ans.begin(); t != ans.end(); t++)
    cout << *t << " ";
  return 0;
}

```