#include <bits/stdc++.h>
using namespace std;
long long trailing_zeros(long long n) {
  long long cnt = 0, f = 5;
  while (f <= n) {
    cnt += (n / f);
    f *= 5;
  }
  return cnt;
}
int main() {
  long long n;
  cin >> n;
  long long high = 5000000000000000000, low = 0, mid, res = -1, temp, zero;
  while (low < high) {
    mid = (low + high) >> 1;
    zero = trailing_zeros(mid);
    if (zero == n) {
      res = mid;
      break;
    }
    if (zero > n)
      high = mid;
    else
      low = mid + 1;
  }
  if (res == -1) {
    cout << 0;
  } else {
    res = res - (res % 5);
    cout << 5 << endl;
    for (int i = 0; i < 5; i++) cout << res + i << " ";
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long trailing_zeros(long long n) {
  long long cnt = 0, f = 5;
  while (f <= n) {
    cnt += (n / f);
    f *= 5;
  }
  return cnt;
}
int main() {
  long long n;
  cin >> n;
  long long high = 5000000000000000000, low = 0, mid, res = -1, temp, zero;
  while (low < high) {
    mid = (low + high) >> 1;
    zero = trailing_zeros(mid);
    if (zero == n) {
      res = mid;
      break;
    }
    if (zero > n)
      high = mid;
    else
      low = mid + 1;
  }
  if (res == -1) {
    cout << 0;
  } else {
    res = res - (res % 5);
    cout << 5 << endl;
    for (int i = 0; i < 5; i++) cout << res + i << " ";
  }
  return 0;
}

```