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;
int k, n;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long i = 0;
  do {
    i += 5;
    for (long long j = i; !(j % 5); j /= 5) k++;
  } while (k < n);
  if (k == n)
    cout << 5 << endl
         << i << " " << i + 1 << " " << i + 2 << " " << i + 3 << " " << i + 4
         << endl;
  else
    cout << 0 << endl;
  return 0;
}