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 <class T1, class T2>
bool chmin(T1 &a, T2 b) {
  return b < a && (a = b, true);
}
template <class T1, class T2>
bool chmax(T1 &a, T2 b) {
  return a < b && (a = b, true);
}
int main() {
  int m;
  cin >> m;
  int two = 0;
  int five = 0;
  vector<int> ans;
  for (int i = 1;; i++) {
    int t = i;
    while (t % 2 == 0) two++, t /= 2;
    while (t % 5 == 0) five++, t /= 5;
    if (min(two, five) == m) ans.push_back(i);
    if (min(two, five) > m) break;
  }
  cout << ans.size() << endl;
  for (int i = 0; i < (ans.size()); i++) printf("%d ", ans[i]);
  cout << endl;
  return 0;
}