#include <bits/stdc++.h>
using namespace std;
vector<long long> z;
void pr() {
  long long i, x, y, s;
  for (i = 1; i <= 500000; i++) {
    x = 1;
    y = 5;
    s = 0;
    while (x) {
      x = i / y;
      s += x;
      y *= 5;
    }
    z.push_back(s);
  }
}
int main() {
  long long n, i, s = 0;
  cin >> n;
  pr();
  vector<long long> p;
  for (i = 0; i < 500000; i++) {
    if (z[i] == n) {
      s++;
      p.push_back(i + 1);
    }
  }
  cout << s << "\n";
  if (s) {
    for (i = 0; i < s; ++i) cout << p[i] << " ";
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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<long long> z;
void pr() {
  long long i, x, y, s;
  for (i = 1; i <= 500000; i++) {
    x = 1;
    y = 5;
    s = 0;
    while (x) {
      x = i / y;
      s += x;
      y *= 5;
    }
    z.push_back(s);
  }
}
int main() {
  long long n, i, s = 0;
  cin >> n;
  pr();
  vector<long long> p;
  for (i = 0; i < 500000; i++) {
    if (z[i] == n) {
      s++;
      p.push_back(i + 1);
    }
  }
  cout << s << "\n";
  if (s) {
    for (i = 0; i < s; ++i) cout << p[i] << " ";
  }
  return 0;
}

```