#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1E9 + 7, INF = 2E18 + 5;
const double PI = 2 * acos(0.0);
const long double EPS = 1.0E-14;
vector<long long> vec;
long long trailingZero(long long n) {
  long long zero = 0;
  while (n / 5 != 0) {
    zero += (n / 5);
    n /= 5;
  }
  return zero;
}
int main() {
  long long n, L, R, mid, zero, ans = -1;
  cin >> n;
  L = 1, R = 5 * n;
  while (L <= R) {
    mid = (L + R) / 2;
    zero = trailingZero(mid);
    if (zero == n) {
      ans = mid;
      R = mid - 1;
    } else if (zero > n) {
      R = mid - 1;
    } else if (zero < n) {
      L = mid + 1;
    }
  }
  if (ans == -1)
    cout << 0;
  else {
    for (int i = ans;; i++) {
      if (trailingZero(i) == n) {
        vec.push_back(i);
      } else
        break;
    }
    cout << vec.size() << "\n";
    for (int i = 0; i < vec.size(); i++) {
      cout << vec[i] << " ";
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Please formulate 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;
const long long MOD = 1E9 + 7, INF = 2E18 + 5;
const double PI = 2 * acos(0.0);
const long double EPS = 1.0E-14;
vector<long long> vec;
long long trailingZero(long long n) {
  long long zero = 0;
  while (n / 5 != 0) {
    zero += (n / 5);
    n /= 5;
  }
  return zero;
}
int main() {
  long long n, L, R, mid, zero, ans = -1;
  cin >> n;
  L = 1, R = 5 * n;
  while (L <= R) {
    mid = (L + R) / 2;
    zero = trailingZero(mid);
    if (zero == n) {
      ans = mid;
      R = mid - 1;
    } else if (zero > n) {
      R = mid - 1;
    } else if (zero < n) {
      L = mid + 1;
    }
  }
  if (ans == -1)
    cout << 0;
  else {
    for (int i = ans;; i++) {
      if (trailingZero(i) == n) {
        vec.push_back(i);
      } else
        break;
    }
    cout << vec.size() << "\n";
    for (int i = 0; i < vec.size(); i++) {
      cout << vec[i] << " ";
    }
    cout << endl;
  }
  return 0;
}

```