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;
vector<unsigned long long int> v1;
multimap<unsigned long long int, unsigned long long int> m;
unsigned long long int i, j, n, p, t, ct = 0, sum = 0, k, g = 0;
set<unsigned long long int> s, s1;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  for (j = 0; j <= 400010; j++) {
    ct = 0;
    for (i = 5; j / i >= 1; i *= 5) {
      ct += j / i;
    }
    m.insert(pair<unsigned long long int, unsigned long long int>(ct, j));
  }
  cin >> t;
  if (m.find(t) == m.end())
    cout << "0";
  else {
    cout << "5" << endl;
    auto i1 = m.lower_bound(t);
    auto i2 = m.upper_bound(t);
    while (i1 != i2) {
      if (i1->first == t) cout << i1->second << " ";
      i1++;
    }
  }
}