#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e9 + 9;
const int N = 1e5 + 10;
int m;
vector<int> f;
void init() { cin >> m; }
void work() {
  int cnt = 0, now;
  for (int i = 1; i <= 5 * m + 10 && cnt <= m; i++) {
    now = i;
    while (now % 5 == 0) cnt++, now /= 5;
    if (cnt == m) f.push_back(i);
  }
  cout << f.size() << endl;
  if (f.size() > 0) {
    for (int i = 0; i < f.size(); i++) cout << f[i] << " ";
  }
  cout << endl;
}
int main() {
  init();
  work();
  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;
const long long MOD = 1e9 + 7;
const long long INF = 1e9 + 9;
const int N = 1e5 + 10;
int m;
vector<int> f;
void init() { cin >> m; }
void work() {
  int cnt = 0, now;
  for (int i = 1; i <= 5 * m + 10 && cnt <= m; i++) {
    now = i;
    while (now % 5 == 0) cnt++, now /= 5;
    if (cnt == m) f.push_back(i);
  }
  cout << f.size() << endl;
  if (f.size() > 0) {
    for (int i = 0; i < f.size(); i++) cout << f[i] << " ";
  }
  cout << endl;
}
int main() {
  init();
  work();
  return 0;
}

```