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 int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
vector<long long int> tokenize(string line) {
  vector<long long int> n;
  string s = "";
  for (int i = 0; i < line.length(); ++i) {
    if (line[i] == ' ') {
      long long int x = stoi(s);
      n.push_back(x);
      s = "";
    } else {
      s += line[i];
    }
  }
  long long int x = stoi(s);
  n.push_back(x);
  return n;
}
long long int trailingzero(long long int n) {
  long long int count = 0;
  for (long long int i = 5; n / i >= 1; i *= 5) {
    count += n / i;
  }
  return count;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  unsigned long long int n;
  cin >> n;
  map<int, vector<int> > list;
  for (int i = 1; i <= 400020; ++i) {
    long long int c = trailingzero(i);
    list[c].push_back(i);
  }
  auto it = list.find(n);
  if (it == list.end()) {
    cout << 0 << "\n";
  } else {
    vector<int> v = (*it).second;
    cout << v.size() << "\n";
    for (int x : v) {
      cout << x << " ";
    }
    cout << ""
         << "\n";
  }
  return 0;
}