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 count_zero(int n) {
  int result = 0;
  while (n) {
    result += n / 5;
    n = n / 5;
  }
  return result;
}
int solve(int m) {
  int flag = 0;
  for (int i = 0; i < 1000000; i++) {
    if (count_zero(i) == m) {
      cout << "5" << endl;
      cout << i << " " << i + 1 << " " << i + 2 << " " << i + 3 << " " << i + 4
           << " ";
      flag = 1;
      break;
    }
  }
  if (flag == 1) {
    return 0;
  } else {
    return 1;
  }
}
int main() {
  int m;
  cin >> m;
  int flag = solve(m);
  if (flag == 1) {
    cout << "0";
  }
}