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;
vector<int> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int a, b, c, x, y, z, i, j, k, m;
  cin >> m;
  for (i = 1; i <= 1000000; ++i) {
    x = 0;
    z = 5;
    while (z <= i) {
      x += (i / z);
      z *= 5;
    }
    if (x == m) v.push_back(i);
    if (x > m) break;
  }
  cout << v.size() << endl;
  if (v.size()) {
    for (i = 0; i < v.size(); ++i) {
      cout << v[i] << " ";
    }
    cout << endl;
  }
  return 0;
}