#include <bits/stdc++.h>
using namespace std;
int m;
int cnt;
vector<int> ans;
int main() {
  cin >> m;
  for (int i = 1;; ++i) {
    int c2 = 0, c5 = 0;
    int t = i;
    while (t > 0) {
      c2 += t / 2;
      t /= 2;
    }
    t = i;
    while (t > 0) {
      c5 += t / 5;
      t /= 5;
    }
    int num = min(c2, c5);
    if (num > m) break;
    if (num == m) {
      ++cnt;
      ans.push_back(i);
    }
  }
  cout << cnt << endl;
  for (int i = 0; i < (ans.size()); i++)
    cout << ans[i] << (i < cnt - 1 ? ' ' : '\n');
  return 0;
}

### Prompt

Generate a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int m;
int cnt;
vector<int> ans;
int main() {
  cin >> m;
  for (int i = 1;; ++i) {
    int c2 = 0, c5 = 0;
    int t = i;
    while (t > 0) {
      c2 += t / 2;
      t /= 2;
    }
    t = i;
    while (t > 0) {
      c5 += t / 5;
      t /= 5;
    }
    int num = min(c2, c5);
    if (num > m) break;
    if (num == m) {
      ++cnt;
      ans.push_back(i);
    }
  }
  cout << cnt << endl;
  for (int i = 0; i < (ans.size()); i++)
    cout << ans[i] << (i < cnt - 1 ? ' ' : '\n');
  return 0;
}

```