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;
long a[100001];
int main() {
  long m, temp;
  cin >> m;
  a[1] = 5;
  long k = 2;
  long j = 2;
  while (j <= 100000) {
    if (k % 5 == 0) {
      long t = k;
      long c = a[j - 1];
      while (t % 5 == 0) {
        a[j++] = 0;
        t /= 5;
      }
      a[j] = c + 5;
    } else {
      a[j] = a[j - 1] + 5;
    }
    j++;
    k++;
  }
  if (a[m] == 0) {
    cout << 0 << endl;
    return 0;
  }
  cout << 5 << endl;
  for (long i = a[m]; i <= a[m] + 4; i++) {
    cout << i << " ";
  }
  cout << endl;
  return 0;
}