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;
int m;
long long sum;
int numOfZero(int n) {
  int num = 0, i;
  for (i = 5; i <= n; i *= 5) {
    num += n / i;
  }
  return num;
}
int main() {
  ios::sync_with_stdio(false);
  while (cin >> m) {
    queue<int> q;
    sum = 0;
    for (int i = 1;; i++) {
      if (numOfZero(i) == m) {
        q.push(i);
        sum++;
      } else if (numOfZero(i) > m)
        break;
    }
    if (sum) {
      cout << sum << endl;
      cout << q.front();
      q.pop();
      while (!q.empty()) {
        cout << " " << q.front();
        q.pop();
      }
      cout << endl;
    } else
      cout << "0" << endl;
  }
  return 0;
}