#include <bits/stdc++.h>
using namespace std;
const int INF = 10000000;
const int mod = 1000000007;
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  vector<long long> v;
  for (long long i = 5; i <= 999999; i++) {
    int cnt = 0;
    for (long long j = 1;; j++) {
      if (i / power(5, j) > 0) {
        cnt += (i / power(5, j));
      }
      if (i / power(5, j) <= 0) {
        break;
      }
    }
    if (cnt == n) {
      v.push_back(i);
    }
  }
  cout << v.size() << '\n';
  for (auto q : v) {
    cout << q << " ";
  }
  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 int INF = 10000000;
const int mod = 1000000007;
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  vector<long long> v;
  for (long long i = 5; i <= 999999; i++) {
    int cnt = 0;
    for (long long j = 1;; j++) {
      if (i / power(5, j) > 0) {
        cnt += (i / power(5, j));
      }
      if (i / power(5, j) <= 0) {
        break;
      }
    }
    if (cnt == n) {
      v.push_back(i);
    }
  }
  cout << v.size() << '\n';
  for (auto q : v) {
    cout << q << " ";
  }
  return 0;
}

```