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() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout.setf(ios_base::fixed);
  cout.precision(28);
  int t = 0, f = 0, i = 1;
  int m;
  cin >> m;
  vector<int> ans;
  while (true) {
    int z = i;
    while (z % 2 == 0) {
      ++t;
      z /= 2;
    }
    while (z % 5 == 0) {
      ++f;
      z /= 5;
    }
    int tn = min(t, f);
    if (tn > m) {
      break;
    }
    if (tn == m) {
      ans.push_back(i);
    }
    ++i;
  }
  cout << ans.size() << endl;
  for (auto x : ans) {
    cout << x << " ";
  }
  return 0;
}