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 no_of_zeroes(long long n) {
  long ans = 0;
  long a = 5;
  while (true) {
    ans += n / a;
    if (n / a == 0) break;
    a *= 5;
  }
  return ans;
}
vector<long long> answer;
int main() {
  ios::sync_with_stdio(false);
  long long m, n;
  long low = 1, high = 1e9;
  long mid;
  cin >> m;
  while (low < high) {
    long mid = (low + high) >> 1;
    if (no_of_zeroes(mid) < m) {
      low = mid + 1;
    } else {
      high = mid;
    }
  }
  while (no_of_zeroes(low) == m) {
    answer.push_back(low);
    ++low;
  }
  cout << (int)answer.size() << '\n';
  for (int i = 0; i < (int)answer.size(); ++i) cout << answer[i] << ' ';
}