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>
std::vector<int> a;
int n, c2, c5;
int main() {
  scanf("%d", &n);
  int cnt = 0;
  int cur = 0;
  while (true) {
    ++cur;
    int m = cur;
    while (m % 2 == 0) {
      m /= 2;
      ++c2;
    }
    while (m % 5 == 0) {
      m /= 5;
      ++c5;
    }
    if (c2 < c5) {
      cnt += c2;
      c2 = 0;
      c5 -= c2;
    } else {
      cnt += c5;
      c5 = 0;
      c2 -= c5;
    }
    if (cnt == n) a.push_back(cur);
    if (cnt > n) break;
  }
  printf("%d\n", (int)a.size());
  for (auto e : a) printf("%d ", e);
  printf("\n");
  return 0;
}