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 MN = 600000;
long long m;
long long count(int n) {
  long long ans = 0;
  for (int i = 5; n / i >= 1; i *= 5) {
    ans += n / i;
  }
  return ans;
}
int main(void) {
  cin >> m;
  if (m == 5) {
    cout << "0\n";
  } else {
    long long ans = 0;
    vector<long long> num;
    for (long long i = 5; i < MN; i++) {
      if (count(i) > m) break;
      if (count(i) == m) {
        ans++;
        num.push_back(i);
      }
    }
    cout << ans << "\n";
    for (int i = 0; i < num.size(); i++) {
      cout << num[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}