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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = ((ans % m) * (a % m)) % m;
    b /= 2;
    a = ((a % m) * (a % m)) % m;
  }
  return ans % m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long m, n;
  cin >> m;
  long long ans = 0;
  long long fans = -1;
  for (int i = 0; i < 1000000; i++) {
    n = i;
    ans = 0;
    while (n > 0) {
      ans += (n / 5);
      n /= 5;
    }
    if (ans == m) {
      fans = i;
      break;
    }
  }
  if (fans == -1) {
    cout << "0";
    return 0;
  }
  vector<int> v;
  v.push_back(fans);
  fans++;
  bool two = false, five = false;
  while (true) {
    if (fans % 2 == 0) two = true;
    if (fans % 5 == 0) five = true;
    if (two && five) break;
    v.push_back(fans);
    fans++;
  }
  cout << v.size() << "\n";
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
}