#include <bits/stdc++.h>
using namespace std;
long long i64 = 4611686018427387904;
long double dist(long double x1, long double y1, long double x2,
                 long double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long nok(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int cc = 0;
  long long ch = 5;
  vector<long long> rz;
  while (cc <= n) {
    long long k = ch;
    while (k % 5 == 0) cc++, k /= 5;
    if (cc == n) {
      rz.push_back(ch);
      ch++;
    } else if (cc < n) {
      ch++;
    } else if (cc > n) {
      break;
    }
  }
  cout << rz.size() << endl;
  for (int i = 0; i < rz.size(); i++) cout << rz[i] << " ";
}

### Prompt

Generate 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;
long long i64 = 4611686018427387904;
long double dist(long double x1, long double y1, long double x2,
                 long double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long nok(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int cc = 0;
  long long ch = 5;
  vector<long long> rz;
  while (cc <= n) {
    long long k = ch;
    while (k % 5 == 0) cc++, k /= 5;
    if (cc == n) {
      rz.push_back(ch);
      ch++;
    } else if (cc < n) {
      ch++;
    } else if (cc > n) {
      break;
    }
  }
  cout << rz.size() << endl;
  for (int i = 0; i < rz.size(); i++) cout << rz[i] << " ";
}

```