#include <bits/stdc++.h>
using namespace std;
int n, m, f;
vector<int> ans;
int zz(int k) {
  int z = 0;
  while (k) {
    k /= 5;
    z += k;
  }
  return z;
}
int main() {
  cin >> n;
  int low = 0, hight = (int)1e9;
  while (low < hight) {
    int md = (low + hight) >> 1;
    if (zz(md) < n)
      low = md + 1;
    else
      hight = md;
  }
  while (zz(low) == n) {
    ans.push_back(low);
    low++;
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[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 n, m, f;
vector<int> ans;
int zz(int k) {
  int z = 0;
  while (k) {
    k /= 5;
    z += k;
  }
  return z;
}
int main() {
  cin >> n;
  int low = 0, hight = (int)1e9;
  while (low < hight) {
    int md = (low + hight) >> 1;
    if (zz(md) < n)
      low = md + 1;
    else
      hight = md;
  }
  while (zz(low) == n) {
    ans.push_back(low);
    low++;
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << ' ';
  return 0;
}

```