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;
const int INF = (int)2e9;
const long long INFL = (long long)9e18;
const int MAXINT = ((~0) ^ (1 << 31));
const long long MAXLL = ((~0) ^ ((long long)1 << 63));
template <class T>
inline T pow2(T a) {
  return a * a;
}
template <class T>
inline bool mineq(T& a, T b) {
  return (a > b) ? (a = b, true) : false;
}
template <class T>
inline bool maxeq(T& a, T b) {
  return (a < b) ? (a = b, true) : false;
}
int search(int m) {
  int cnt = 0;
  for (int i = 1; i < (int)(1000000); i++) {
    int tmp = i;
    while (tmp % 5 == 0) {
      tmp /= 5;
      m--;
    }
    if (m <= 0) return i;
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int m;
  cin >> m;
  int l = search(m), r = search(m + 1);
  cout << r - l << endl;
  for (int i = l; i < (int)(r); i++) cout << i << " ";
  cout << endl;
}