#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1000 * 1000 * 1000 + 7;
const int INF = 2000 * 1000 * 1000;
template <typename T>
inline T sqr(T n) {
  return n * n;
}
int m;
vector<int> ans;
int main() {
  scanf("%d", &m);
  int cnt = 0;
  for (int i = 1;; i++) {
    int temp = i;
    while (temp % 5 == 0) {
      cnt++;
      temp /= 5;
    }
    if (cnt == m) {
      ans.push_back(i);
    } else if (cnt > m) {
      break;
    }
  }
  printf("%llu\n", ans.size());
  for (int x : ans) {
    printf("%d ", x);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1000 * 1000 * 1000 + 7;
const int INF = 2000 * 1000 * 1000;
template <typename T>
inline T sqr(T n) {
  return n * n;
}
int m;
vector<int> ans;
int main() {
  scanf("%d", &m);
  int cnt = 0;
  for (int i = 1;; i++) {
    int temp = i;
    while (temp % 5 == 0) {
      cnt++;
      temp /= 5;
    }
    if (cnt == m) {
      ans.push_back(i);
    } else if (cnt > m) {
      break;
    }
  }
  printf("%llu\n", ans.size());
  for (int x : ans) {
    printf("%d ", x);
  }
  return 0;
}

```