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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int m;
  cin >> m;
  vector<int> v;
  int s2 = 0;
  int s5 = 0;
  for (int i = 1;; i++) {
    int x = i;
    while (x % 2 == 0) {
      s2++;
      x /= 2;
    }
    while (x % 5 == 0) {
      s5++;
      x /= 5;
    }
    if (min(s2, s5) == m) {
      v.push_back(i);
    }
    if (min(s2, s5) > m) break;
  }
  cout << v.size() << endl;
  for (int vv : v) {
    cout << vv << " ";
  }
  cout << endl;
}