#include <bits/stdc++.h>
using namespace std;
const int N = 500000;
long long fast_pow(long long base, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * base) % 1000000007;
    base = (base * base) % 1000000007;
    n = n >> 1;
  }
  return res;
}
int pre[N + 5];
int main() {
  for (int i = 5; i <= N; i = i * 5) {
    for (int j = 1; i * j <= N; j++) {
      pre[i * j]++;
    }
  }
  for (int i = 1; i <= N; i++) {
    pre[i] += pre[i - 1];
  }
  int m;
  cin >> m;
  vector<int> ans;
  for (int i = 1; i <= N; i++) {
    if (pre[i] == m) {
      ans.push_back(i);
    }
  }
  int c = ans.size();
  sort(ans.begin(), ans.end());
  cout << c << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  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 N = 500000;
long long fast_pow(long long base, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * base) % 1000000007;
    base = (base * base) % 1000000007;
    n = n >> 1;
  }
  return res;
}
int pre[N + 5];
int main() {
  for (int i = 5; i <= N; i = i * 5) {
    for (int j = 1; i * j <= N; j++) {
      pre[i * j]++;
    }
  }
  for (int i = 1; i <= N; i++) {
    pre[i] += pre[i - 1];
  }
  int m;
  cin >> m;
  vector<int> ans;
  for (int i = 1; i <= N; i++) {
    if (pre[i] == m) {
      ans.push_back(i);
    }
  }
  int c = ans.size();
  sort(ans.begin(), ans.end());
  cout << c << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  return 0;
}

```