#include <bits/stdc++.h>
using namespace std;
int main() {
  int m, i;
  scanf("%d", &m);
  vector<int> ans;
  for (i = 1; i <= 1000000; ++i) {
    int cnt = 0, fact = 5;
    while (fact <= i) {
      cnt += i / fact, fact *= 5;
    }
    if (cnt == m) ans.push_back(i);
  }
  int sz = ans.size();
  printf("%d\n", sz);
  for (i = 0; i < sz; ++i) {
    if (i) putchar(' ');
    printf("%d", ans[i]);
  }
  puts("");
  return 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;
int main() {
  int m, i;
  scanf("%d", &m);
  vector<int> ans;
  for (i = 1; i <= 1000000; ++i) {
    int cnt = 0, fact = 5;
    while (fact <= i) {
      cnt += i / fact, fact *= 5;
    }
    if (cnt == m) ans.push_back(i);
  }
  int sz = ans.size();
  printf("%d\n", sz);
  for (i = 0; i < sz; ++i) {
    if (i) putchar(' ');
    printf("%d", ans[i]);
  }
  puts("");
  return 0;
}

```