#include <bits/stdc++.h>
using namespace std;
int m[100005];
int main() {
  long long int m, f, i, c, j;
  cin >> m;
  f = 0;
  for (i = 0;; i++) {
    c = 0;
    for (j = 5; i / j >= 1; j = j * 5) {
      c = c + i / j;
    }
    if (c == m) {
      f = 1;
      break;
    } else if (c > m) {
      f = 2;
      break;
    }
  }
  if (f == 1) {
    cout << 5 << endl;
    for (j = i; j <= i + 4; j++) {
      cout << j << " ";
    }
    cout << endl;
  } else if (f == 2) {
    cout << "0" << endl;
  }
  return 0;
}

### Prompt

Generate 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 m[100005];
int main() {
  long long int m, f, i, c, j;
  cin >> m;
  f = 0;
  for (i = 0;; i++) {
    c = 0;
    for (j = 5; i / j >= 1; j = j * 5) {
      c = c + i / j;
    }
    if (c == m) {
      f = 1;
      break;
    } else if (c > m) {
      f = 2;
      break;
    }
  }
  if (f == 1) {
    cout << 5 << endl;
    for (j = i; j <= i + 4; j++) {
      cout << j << " ";
    }
    cout << endl;
  } else if (f == 2) {
    cout << "0" << endl;
  }
  return 0;
}

```