#include <bits/stdc++.h>
using namespace std;
int a[101000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int x = 0;
  for (int i = 1; i < 100100; ++i) {
    x += 5;
    int add = 0;
    int copy = x;
    while (copy % 5 == 0) {
      copy /= 5;
      add++;
    }
    i += add - 1;
    a[i] = x;
  }
  x = a[n];
  if (a[n] != 0) {
    cout << 5 << endl;
    for (int i = 0; i < 5; ++i) {
      cout << x << " ";
      x++;
    }
  } else
    cout << 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 a[101000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int x = 0;
  for (int i = 1; i < 100100; ++i) {
    x += 5;
    int add = 0;
    int copy = x;
    while (copy % 5 == 0) {
      copy /= 5;
      add++;
    }
    i += add - 1;
    a[i] = x;
  }
  x = a[n];
  if (a[n] != 0) {
    cout << 5 << endl;
    for (int i = 0; i < 5; ++i) {
      cout << x << " ";
      x++;
    }
  } else
    cout << 0;
}

```