#include <bits/stdc++.h>
using namespace std;
vector<int> v[100001];
int mod = 1000003;
int main() {
  long long n, k = 0, a[3001] = {0}, x, y;
  cin >> n;
  x = 5;
  while (n > 0) {
    y = x;
    k = 0;
    while (y % 5 == 0) {
      y = y / 5;
      k++;
    }
    n = n - k;
    x = x + 5;
  }
  if (n == 0)
    cout << 5 << endl
         << x - 5 << ' ' << x - 4 << ' ' << x - 3 << ' ' << x - 2 << ' '
         << x - 1;
  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;
vector<int> v[100001];
int mod = 1000003;
int main() {
  long long n, k = 0, a[3001] = {0}, x, y;
  cin >> n;
  x = 5;
  while (n > 0) {
    y = x;
    k = 0;
    while (y % 5 == 0) {
      y = y / 5;
      k++;
    }
    n = n - k;
    x = x + 5;
  }
  if (n == 0)
    cout << 5 << endl
         << x - 5 << ' ' << x - 4 << ' ' << x - 3 << ' ' << x - 2 << ' '
         << x - 1;
  else
    cout << 0;
}

```