#include <bits/stdc++.h>
using namespace std;
void prepere_streams() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
inline int count(int x) {
  int res = 0;
  while (x) {
    res += x / 5;
    x /= 5;
  }
  return res;
}
int main() {
  int m;
  cin >> m;
  int l = 1;
  int r = 500000;
  int me;
  int q;
  int h;
  while (r - l > 1) {
    me = (l + r) / 2;
    q = count(me);
    if (q > m) r = me;
    if (q < m) l = me;
    if (q == m) {
      h = me;
      while (count(h) == m) ++h;
      while (count(me) == m) --me;
      cout << h - me - 1 << '\n';
      for (int i = me + 1; i < h; ++i) cout << i << ' ';
      cin >> m;
      return 0;
    }
  }
  cout << ((count(r) == m) + (count(l) == m)) << '\n';
  if ((count(l) == m)) cout << l << ' ';
  if (count(r) == m) cout << r;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void prepere_streams() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
inline int count(int x) {
  int res = 0;
  while (x) {
    res += x / 5;
    x /= 5;
  }
  return res;
}
int main() {
  int m;
  cin >> m;
  int l = 1;
  int r = 500000;
  int me;
  int q;
  int h;
  while (r - l > 1) {
    me = (l + r) / 2;
    q = count(me);
    if (q > m) r = me;
    if (q < m) l = me;
    if (q == m) {
      h = me;
      while (count(h) == m) ++h;
      while (count(me) == m) --me;
      cout << h - me - 1 << '\n';
      for (int i = me + 1; i < h; ++i) cout << i << ' ';
      cin >> m;
      return 0;
    }
  }
  cout << ((count(r) == m) + (count(l) == m)) << '\n';
  if ((count(l) == m)) cout << l << ' ';
  if (count(r) == m) cout << r;
}

```