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 get(int n) {
  int z = 0;
  int i = 5;
  while ((n / i) > 0) {
    z += (n / i);
    i *= 5;
  }
  return z;
}
int main() {
  int m;
  int i;
  cin >> m;
  int low;
  int high;
  low = 1;
  high = (int)1e9;
  while (low < high) {
    int mid;
    mid = (low + high) >> 1;
    int z = get(mid);
    if (z < m) {
      low = mid + 1;
    } else {
      high = mid;
    }
  }
  if (get(low) == m) {
    puts("5");
    for (i = low; i < low + 5; i++) cout << i << " ";
    puts(" ");
  } else
    cout << "0\n";
  return 0;
}