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;
long fact(long a) {
  long i = 1, ans = 0;
  while (a / pow(5, i)) {
    ans += a / pow(5, i);
    i++;
  }
  return ans;
}
int main() {
  long m;
  cin >> m;
  if (m == 0) {
    cout << "4\n1 2 3 4";
    return 0;
  }
  long low = 1;
  long high = 5 * (m + 1);
  long mid = (low + high) / 2;
  while (low < high) {
    mid = (low + high) / 2;
    if (fact(mid) == m) {
      long a = long(mid / 5) * 5;
      cout << 5 << "\n";
      for (__typeof(5) i = 0; i < 5; i++) cout << a + i << " ";
      return 0;
    } else if (fact(mid) < m)
      low = mid + 1;
    else
      high = mid;
  }
  cout << 0;
  return 0;
}