#include <bits/stdc++.h>
using namespace std;
int n;
int ans[100010], cnt;
int cal(int x) {
  int ret = 0;
  while (x % 5 == 0) {
    x /= 5;
    ret++;
  }
  return ret;
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  int tmp2 = 0;
  int tmp5 = 0;
  for (i = 1; i <= 1000000; i++) {
    if (i % 5 == 0) tmp5 += cal(i);
    if (tmp5 == n) ans[++cnt] = i;
  }
  cout << cnt << endl;
  for (i = 1; i <= cnt; i++) {
    cout << ans[i] << " ";
  }
  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 n;
int ans[100010], cnt;
int cal(int x) {
  int ret = 0;
  while (x % 5 == 0) {
    x /= 5;
    ret++;
  }
  return ret;
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  int tmp2 = 0;
  int tmp5 = 0;
  for (i = 1; i <= 1000000; i++) {
    if (i % 5 == 0) tmp5 += cal(i);
    if (tmp5 == n) ans[++cnt] = i;
  }
  cout << cnt << endl;
  for (i = 1; i <= cnt; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

```