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 t = 1;
void solve() {
  long long n, m, c = 0, d = 0;
  vector<long long> v;
  cin >> m;
  for (long long i = 1; i <= 1000001; i++) {
    long long a = i;
    while (a % 5 == 0) {
      a = a / 5;
      d++;
    }
    if (d == m) {
      v.push_back(i);
      c++;
    }
    if (d > m) break;
  }
  cout << c << "\n";
  if (c > 0) {
    for (auto e : v) cout << e << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  while (t--) {
    solve();
  }
}