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 trailingZeros(int n) {
  int count = 0;
  for (int i = 5; n / i >= 1; i *= 5) count += n / i;
  return count;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m;
  cin >> m;
  int l = 0, u = int(1e9);
  while (l < u) {
    int mid = l + (u - l) / 2;
    if (trailingZeros(mid) < m)
      l = mid + 1;
    else
      u = mid;
  }
  int lb = l;
  l = 0, u = int(1e9);
  while (l < u) {
    int mid = l + (u - l) / 2 + 1;
    if (m < trailingZeros(mid))
      u = mid - 1;
    else
      l = mid;
  }
  int ub = l;
  if (ub < lb) {
    cout << 0;
    return 0;
  }
  cout << ub - lb + 1 << "\n";
  for (int i = lb; i <= ub; i++) cout << i << " ";
  return 0;
}