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() {
  long long int a, b, c, t, i, j, k, l, m, n, o, p;
  cin >> n;
  p = 0;
  for (i = 5;; i = i + 5) {
    o = 1;
    p = 0;
    m = 5;
    while (o == 1) {
      j = i / m;
      p += j;
      m *= 5;
      if (j == 0) break;
    }
    if (p == n) {
      cout << "5\n";
      for (k = i; k < i + 5; k++) cout << k << " ";
      break;
    }
    if (p > n) {
      cout << "0\n";
      break;
    }
  }
}