#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> ret;
pair<int, int> f(int a) {
  int q = 0, w = 0;
  while (a % 2 == 0) {
    ++q;
    a /= 2;
  }
  while (a % 5 == 0) {
    ++w;
    a /= 5;
  }
  return make_pair(q, w);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  pair<int, int> last{0, 0};
  for (int a = 1;; ++a) {
    pair<int, int> r = last;
    pair<int, int> p = f(a);
    r.first += p.first;
    r.second += p.second;
    if (min(r.first, r.second) == n) {
      ret.push_back(a);
    } else if (min(r.first, r.second) > n) {
      break;
    }
    last = r;
  }
  cout << ret.size() << "\n";
  for (int x : ret) {
    cout << x << " ";
  }
  return 0;
}

### Prompt

Your task is to create 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;
int n;
vector<int> ret;
pair<int, int> f(int a) {
  int q = 0, w = 0;
  while (a % 2 == 0) {
    ++q;
    a /= 2;
  }
  while (a % 5 == 0) {
    ++w;
    a /= 5;
  }
  return make_pair(q, w);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  pair<int, int> last{0, 0};
  for (int a = 1;; ++a) {
    pair<int, int> r = last;
    pair<int, int> p = f(a);
    r.first += p.first;
    r.second += p.second;
    if (min(r.first, r.second) == n) {
      ret.push_back(a);
    } else if (min(r.first, r.second) > n) {
      break;
    }
    last = r;
  }
  cout << ret.size() << "\n";
  for (int x : ret) {
    cout << x << " ";
  }
  return 0;
}

```