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;
double PI = acos(-1);
long long arr[1000005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m;
  cin >> m;
  vector<int> ans;
  for (int i = 0; i <= 1000000; i++) {
    int cnt = 0, k = i;
    while (k % 5 == 0 && k > 0) {
      cnt++;
      k /= 5;
    }
    arr[i] = cnt;
  }
  for (int i = 1; i <= 1000000; i++) arr[i] += arr[i - 1];
  for (int i = 0; i <= 1000000; i++) {
    if (arr[i] == m) ans.push_back(i);
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  return 0;
}