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;
template <typename X>
inline X abs(const X& a) {
  return a < 0 ? -a : a;
}
template <typename X>
inline X sqr(const X& a) {
  return a * a;
}
long long combine(long long n, int k) {
  long long ans = 1;
  for (int i = 0; i < k; i++) ans = (ans * (n - i)) / (i + 1);
  return ans;
}
long long pw(long long a, long long p) {
  if (p == 0) return 1;
  if (p == 1) return a;
  long long half = pw(a, p / 2);
  if (p % 2 == 0) return half * half;
  return half * half * a;
}
long long dis(pair<long long, long long> a, pair<long long, long long> b) {
  return (a.first - b.first) * (a.first - b.first) +
         (a.second - b.second) * (a.second - b.second);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int m;
  cin >> m;
  long long cnt = 0, k = 0;
  int i = 1, j;
  while (cnt < m) {
    j = i;
    if (j % 5 == 0) k++;
    while (j % 5 == 0) cnt++, j /= 5;
    i++;
  }
  if (cnt != m || m == 0)
    cout << 0 << endl;
  else {
    cout << 5 << endl;
    k *= 5;
    cnt = k + 5;
    for (; k < cnt; k++) {
      cout << k << " ";
    }
    cout << endl;
  }
  return 0;
}