#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > cnt;
int fin(int n) {
  int count = 0;
  for (int i = 5; n / i >= 1; i *= 5) count += n / i;
  return count;
}
void init() {
  const int MAX = 1e6;
  cnt.resize(MAX + 1);
  for (int i = 1; i <= 5 * MAX; i++) {
    if (fin(i) <= MAX) cnt[fin(i)].push_back(i);
  }
}
int main() {
  init();
  int m;
  cin >> m;
  cout << cnt[m].size() << "\n";
  for (int i = 0; i < cnt[m].size(); i++) {
    cout << cnt[m][i] << " ";
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
vector<vector<int> > cnt;
int fin(int n) {
  int count = 0;
  for (int i = 5; n / i >= 1; i *= 5) count += n / i;
  return count;
}
void init() {
  const int MAX = 1e6;
  cnt.resize(MAX + 1);
  for (int i = 1; i <= 5 * MAX; i++) {
    if (fin(i) <= MAX) cnt[fin(i)].push_back(i);
  }
}
int main() {
  init();
  int m;
  cin >> m;
  cout << cnt[m].size() << "\n";
  for (int i = 0; i < cnt[m].size(); i++) {
    cout << cnt[m][i] << " ";
  }
  return 0;
}

```