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 <class T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
inline T LCM(T a, T b) {
  if (a < 0) return LCM(-a, b);
  if (b < 0) return LCM(a, -b);
  return a * (b / gcd(a, b));
}
int main() {
  vector<int> ans;
  int n;
  cin >> n;
  int from = 5;
  int c = 0;
  for (int j = 1; j <= 100000; j++) {
    int z = 5 * j;
    while (z % 5 == 0) {
      c++;
      z /= 5;
    }
    if (c == n) {
      for (int k = 5 * j; k < 5 * (j + 1); k++) {
        ans.push_back(k);
      }
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}