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 long z(long long n) {
  long long ans = 0;
  while (n / 5 != 0) {
    ans += n / 5;
    n /= 5;
  }
  return ans;
}
vector<int> ans;
int main() {
  int m;
  scanf("%d", &m);
  for (int i = 1; i < 1000005; i++) {
    if (z(i) == m) {
      ans.push_back(i);
    }
  }
  int sz = ans.size();
  printf("%d\n", sz);
  if (sz) {
    for (int i = 0; i < ans.size(); i++) {
      printf("%d ", ans[i]);
    }
  }
}