#include <bits/stdc++.h>
using namespace std;
int zeros(int i) {
  int ans = 0, j = 5;
  while (i >= j) ans += i / j, j *= 5;
  return ans;
}
int main() {
  int n, m = -1;
  cin >> n;
  vector<long long> v;
  int low = 2, high = 500000;
  while (low < high) {
    int mid = (low + high) / 2;
    m = zeros(mid);
    if (m >= n)
      high = mid;
    else
      low = mid + 1;
  }
  int i = low;
  while (zeros(i) == n) v.push_back(i++);
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    printf("%I64d ", v[i]);
  }
  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 zeros(int i) {
  int ans = 0, j = 5;
  while (i >= j) ans += i / j, j *= 5;
  return ans;
}
int main() {
  int n, m = -1;
  cin >> n;
  vector<long long> v;
  int low = 2, high = 500000;
  while (low < high) {
    int mid = (low + high) / 2;
    m = zeros(mid);
    if (m >= n)
      high = mid;
    else
      low = mid + 1;
  }
  int i = low;
  while (zeros(i) == n) v.push_back(i++);
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    printf("%I64d ", v[i]);
  }
  return 0;
}

```