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;
bool isPerfectSquare(long long int n) {
  long long int lol = sqrt(n);
  if (lol * lol == n)
    return true;
  else
    return false;
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n, x, i = 1, y;
  cin >> n;
  while (x < n) {
    y = i;
    while (y % 5 == 0) {
      x++;
      y = y / 5;
    }
    i++;
  }
  if (x != n) {
    cout << 0 << '\n';
    return 0;
  }
  i--;
  cout << 5 << '\n';
  cout << i << ' ' << i + 1 << ' ' << i + 2 << ' ' << i + 3 << ' ' << i + 4
       << '\n';
  return 0;
}