#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int main() {
  int m;
  while (~scanf("%d", &m)) {
    vector<int> ans;
    for (int n = 1;; ++n) {
      int r = 0;
      for (int x = n / 5; x > 0; x /= 5) r += x;
      if (r > m) break;
      if (r == m) ans.push_back(n);
    }
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < (int)ans.size(); ++i) {
      if (i != 0) putchar(' ');
      printf("%d", ans[i]);
    }
    puts("");
  }
  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;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int main() {
  int m;
  while (~scanf("%d", &m)) {
    vector<int> ans;
    for (int n = 1;; ++n) {
      int r = 0;
      for (int x = n / 5; x > 0; x /= 5) r += x;
      if (r > m) break;
      if (r == m) ans.push_back(n);
    }
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < (int)ans.size(); ++i) {
      if (i != 0) putchar(' ');
      printf("%d", ans[i]);
    }
    puts("");
  }
  return 0;
}

```